TOC 
Network Working GroupA. Kato
Internet-DraftNTT Software Corporation
Intended status: InformationalM. Kanda
Expires: September 20, 2008Nippon Telegraph and Telephone
 Corporation
 March 19, 2008


Camellia Counter mode and Camellia Counter with CBC Mac mode algorithms
draft-kato-camellia-ctrccm-01

Status of this Memo

By submitting this Internet-Draft, each author represents that any applicable patent or other IPR claims of which he or she is aware have been or will be disclosed, and any of which he or she becomes aware will be disclosed, in accordance with Section 6 of BCP 79.

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This Internet-Draft will expire on September 20, 2008.

Abstract

This document describes the algorithms and test vectors of Camellia block cipher algorithm in Counter (CTR) mode and Counter with Cipher Block Chaining MAC (CCM) Mode. The purpose of this document is to make the Camellia-CTR and Camellia-CCM algorithm conveniently available to the Internet Community.


Table of Contents

1.  Introduction
    1.1.  Terminology
2.  The Camellia Cipher Algorithm
    2.1.  Key Size
    2.2.  Weak Keys
    2.3.  Block Size and Padding
    2.4.  Performance
3.  Modes of Operation
    3.1.  Counter
        3.1.1.  Definitions
        3.1.2.  Camellia-CTR
    3.2.  Counter with CBC-MAC
        3.2.1.  Definitions
        3.2.2.  Two main parameters
        3.2.3.  Inputs
        3.2.4.  Authentication
        3.2.5.  Encryption
        3.2.6.  Output
        3.2.7.  Decryption and Authentication Checking
4.  Test Vectors
    4.1.  Camellia-CTR
    4.2.  Camellia-CCM
5.  Security Considerations
6.  IANA Considerations
7.  Acknowledgments
8.  References
    8.1.  Normative
    8.2.  Informative
§  Authors' Addresses
§  Intellectual Property and Copyright Statements



 TOC 

1.  Introduction

This document describes the use of the Camellia block cipher algorithm in Counter mode and Counter with CBC-MAC Mode.

Camellia is a symmetric cipher with a Feistel structure. Camellia was developed jointly by NTT and Mitsubishi Electric Corporation in 2000. It was designed to withstand all known cryptanalytic attacks, and it has been scrutinized by worldwide cryptographic experts. Camellia is suitable for implementation in software and hardware, offering encryption speed in software and hardware implementations that is comparable to Advanced Encryption Standard (AES) [3] (National Institute of Standards and Technology, “Advanced Encryption Standard (AES),” November 2001.).

Camellia supports 128-bit block size and 128-, 192-, and 256-bit key lengths, i.e., the same interface specifications as the AES. Therefore, it is easy to implement Camellia based algorithms by replacing AES block of AES based algorithms to Camellia block.

Camellia is adopted as IETF and several international standardization organizations. Camellia is already adopted as IPSec [4] (Kato, A., Moriai, S., and M. Kanda, “The Camellia Cipher Algorithm and Its Use With IPsec,” December 2005.), TLS [5] (Moriai, S., Kato, A., and M. Kanda, “Addition of Camellia Cipher Suites to Transport Layer Security (TLS),” July 2005.), S/MIME [6] (Moriai, S. and A. Kato, “Use of the Camellia Encryption Algorithm in Cryptographic Message Syntax (CMS),” January 2004.) and XML [7] (Eastlake, D., “Additional XML Security Uniform Resource Identifiers (URIs),” April 2005.). Camellia is adopted for the one of three ISO/IEC international standard cipher [8] (International Organization for Standardization, “Information technology - Security techniques - Encryption algorithms - Part 3: Block ciphers,” July 2005.) as 128bit block cipher(Camellia, AES and SEED). Camellia was selected as a recommended cryptographic primitive by the EU NESSIE (New European Schemes for Signatures, Integrity and Encryption) project [9] (, “The NESSIE project (New European Schemes for Signatures, Integrity and Encryption),” .) and was included in the list of cryptographic techniques for Japanese e-Government systems that was selected by the Japan CRYPTREC (Cryptography Research and Evaluation Committees) [10] (Information-technology Promotion Agency (IPA), “Cryptography Research and Evaluation Committees,” .).

Since optimized source code is provided by several open source lisences, Camellia is also adopted by several open source projects(Openssl, FreeBSD, Linux and Gran Paradiso).

The algorithm specification and object identifiers are described in [1] (Matsui, M., Nakajima, J., and S. Moriai, “A Description of the Camellia Encryption Algorithm,” April 2004.).

The Camellia homepage contains a wealth of information about Camellia, including detailed specification, security analysis, performance figures, reference implementation, optimized implementation, test vectors, and intellectual property information.
 TOC 

1.1.  Terminology

The key words "MUST", "MUST NOT", "REQUIRED", "SHALL", "SHALL NOT", "SHOULD", "SHOULD NOT", "RECOMMENDED", "MAY", and "OPTIONAL" in this document are to be interpreted as described in [2] (Bradner, S., “Key words for use in RFCs to Indicate Requirement Levels,” March 1997.).


 TOC 

2.  The Camellia Cipher Algorithm

All symmetric block cipher algorithms share common characteristics and variables, including mode, key size, weak keys, block size, and rounds. The following sections contain descriptions of the relevant characteristics of Camellia.

The algorithm specification and object identifiers are described in [1] (Matsui, M., Nakajima, J., and S. Moriai, “A Description of the Camellia Encryption Algorithm,” April 2004.).


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2.1.  Key Size

Camellia supports three key sizes: 128 bits, 192 bits, and 256 bits. The default key size is 128 bits, and all implementations MUST support this key size. Implementations MAY also support key sizes of 192 bits and 256 bits.

Camellia uses a different number of rounds for each of the defined key sizes. When a 128-bit key is used, implementations MUST use 18 rounds. When a 192-bit key is used, implementations MUST use 24 rounds. When a 256-bit key is used, implementations MUST use 24 rounds.


 TOC 

2.2.  Weak Keys

At the time of writing this document there are no known weak keys for Camellia.


 TOC 

2.3.  Block Size and Padding

Camellia uses a block size of sixteen octets (128 bits).

Padding is required by the algorithms to maintain a 16-octet (128-bit) block size. Padding MUST be added such that the data to be encrypted (which includes the ESP Pad Length and Next Header fields) has a length that is a multiple of 16 octets.

Because of the algorithm specific padding requirement, no additional padding is required to ensure that the ciphertext terminates on a 4-octet boundary (i.e. maintaining a 16-octet block size guarantees that the ESP Pad Length and Next Header fields will be right aligned within a 4-octet word). Additional padding MAY be included as long as the 16-octet block size is maintained.


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2.4.  Performance

Performance figures of Camellia are available at http://info.isl.ntt.co.jp/crypt/camellia/. NESSIE project has reported performance of Optimized Implementations independently [9] (, “The NESSIE project (New European Schemes for Signatures, Integrity and Encryption),” .).


 TOC 

3.  Modes of Operation

Camellia Counter mode (Camellia-CTR) and Camellia Counter with CBC-MAC (Camellia-CCM) are discussed in this specification.

Counter mode [11] (Dworkin, M., “Recommendation for Block Cipher Modes of Operation - Methods and Techniques,” November 2001.) behave like stream ciphers, but are constructed based on a block cipher primitive (that is, counter mode operation of a block cipher results in a stream cipher.)

Counter with CBC Mac mode is a generic authenticate-and-encrypt block cipher mode [12] (Whiting, D., Housley, R., and N. Ferguson, “Counter with CBC-MAC (CCM),” September 2003.). In this specification, CCM is used with the Camellia [1] (Matsui, M., Nakajima, J., and S. Moriai, “A Description of the Camellia Encryption Algorithm,” April 2004.) block cipher.


 TOC 

3.1.  Counter


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3.1.1.  Definitions

Camellia(K, X)
Output of the Camellia encryption algorithm under the fresh key K applied to the data block X.
N
Nonce.
PT[n]
n-th plain text block splited by 128-bit unit.
X || Y
Concatenation of two octet strings X and Y.
X XOR Y
Bitwise exclusive-OR of two octet strings X and Y of the same length.

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3.1.2.  Camellia-CTR

Camellia-CTR requires the encryptor to generate a unique per-packet value, and communicate this value to the decryptor. This specification calls this per-packet value an initialization vector (IV). The same IV and key combination MUST NOT be used more than once. The encryptor can generate the IV in any manner that ensures uniqueness. Common approaches to IV generation include incrementing a counter for each packet and linear feedback shift registers (LFSRs).

This specification calls for the use of a nonce for additional protection against precomputation attacks. The nonce value need not be secret. However, the nonce MUST be unpredictable prior to the establishment of the IPsec security association that is making use of Camellia-CTR.

Camellia-CTR has many properties that make it an attractive encryption algorithm for in high-speed networking. Camellia-CTR uses the Camellia block cipher to create a stream cipher. Data is encrypted and decrypted by XORing with the key stream produced by Camellia encrypting sequential counter block values. Camellia-CTR is easy to implement, and Camellia-CTR can be pipelined and parallelized. Camellia-CTR also supports key stream precomputation.

Pipelining is possible because Camellia has multiple rounds (see Section 2 (The Camellia Cipher Algorithm).). A hardware implementation (and some software implementations) can create a pipeline by unwinding the loop implied by this round structure. For example, after a 16-octet block has been input, one round later another 16-octet block can be input, and so on. In Camellia-CTR, these inputs are the sequential counter block values used to generate the key stream.

Multiple independent Camellia encrypt implementations can also be used to improve performance. For example, one could use two Camellia encrypt implementations in parallel, to process a sequence of counter block values, doubling the effective throughput.

The sender can precompute the key stream. Since the key stream does not depend on any data in the packet, the key stream can be precomputed once the nonce and IV are assigned. This precomputation can reduce packet latency. The receiver cannot perform similar precomputation because the IV will not be known before the packet arrives.

When used correctly, Camellia-CTR provides a high level of confidentiality. Unfortunately, Camellia-CTR is easy to use incorrectly. Being a stream cipher, any reuse of the per-packet value, called the IV, with the same nonce and key is catastrophic. An IV collision immediately leaks information about the plaintext in both packets. For this reason, it is inappropriate to use this mode of operation with static keys. Extraordinary measures would be needed to prevent reuse of an IV value with the static key across power cycles. To be safe, implementations MUST use fresh keys with Camellia-CTR.

With Camellia-CTR, it is trivial to use a valid ciphertext to forge other (valid to the decryptor) ciphertexts. Thus, it is equally catastrophic to use Camellia-CTR without a companion authentication function. Implementations MUST use Camellia-CCM such case.

To encrypt a payload with Camellia-CTR, the encryptor partitions the plaintext, PT, into 128-bit blocks. The final block need not be 128 bits; it can be less. 

      PT = PT[1] PT[2] ... PT[n]

Each PT block is XORed with a block of the key stream to generate the ciphertext, CT. The Camellia encryption of each counter block results in 128 bits of key stream. The most significant 96 bits of the counter block are set to the nonce value, which is 32 bits, followed by the per-packet IV value, which is 64 bits. The least significant 32 bits of the counter block are initially set to one. This counter value is incremented by one to generate subsequent counter blocks, each resulting in another 128 bits of key stream. The encryption of n plaintext blocks can be summarized as: 

      CTRBLK := N || IV || ONE
      FOR i := 1 to n-1 DO
        CT[i] := PT[i] XOR Camellia(K, CTRBLK)
        CTRBLK := CTRBLK + 1
      END
      CT[n] := PT[n] XOR TRUNC(Camellia(K, CTRBLK))

The TRUNC() function truncates the output of the Camellia encrypt operation to the same length as the final plaintext block, returning the most significant bits.

Decryption is similar. The decryption of n ciphertext blocks can be summarized as: 


      CTRBLK := N || IV || ONE
      FOR i := 1 to n-1 DO
        PT[i] := CT[i] XOR Camellia(K, CTRBLK)
        CTRBLK := CTRBLK + 1
      END
      PT[n] := CT[n] XOR TRUNC(Camellia(K, CTRBLK))

 TOC 

3.2.  Counter with CBC-MAC


 TOC 

3.2.1.  Definitions

l(X)
Octet length of variable X.
M
Number of octets in authentication field. Valid values of are 4, 6, 8, 10, 12, 14, and 16.
M'
3bit number calculated by (M-2)/2.
L
Number of octets in length field. Valid values are from 2 to 8. This number limits maximam length of message and length of N.
L'
3bit number caluculated by L-1.
K
Camellia key. Valid values of l(K) are 16, 24 and 32.
N
Nonce.
m
Message to authenticate and encrypt. l(m) < 2^(8*L).
AAD
Additional authenticated data. 0 =< AAD < 2^64.

 TOC 

3.2.2.  Two main parameters

For the generic CCM mode there are two parameter choices. The first choice is M, the size of the authentication field. The choice of the value for M involves a trade-off between message expansion and the probability that an attacker can undetectably modify a message. Valid values are 4, 6, 8, 10, 12, 14, and 16 octets. The second choice is L, the size of the length field. This value requires a trade-off between the maximum message size and the size of the Nonce. Different applications require different trade-offs, so L is a parameter. Valid values of L range between 2 octets and 8 octets (the value L=1 is reserved). 

       Name  Description                               Size    Encoding
       ----  ----------------------------------------  ------  --------
       M     Number of octets in authentication field  3 bits  (M-2)/2
       L     Number of octets in length field          3 bits  L-1

 TOC 

3.2.3.  Inputs

To authenticate and encrypt a message the following information is required:

  1. An encryption key K suitable for the block cipher.
  2. A nonce N of 15-L octets. Within the scope of any encryption key K, the nonce value MUST be unique. That is, the set of nonce values used with any given key MUST NOT contain any duplicate values. Using the same nonce for two different messages encrypted with the same key destroys the security properties of this mode.
  3. The message m, consisting of a string of l(m) octets where 0 <= l(m) < 2^(8*L). The length restriction ensures that l(m) can be encoded in a field of L octets.
  4. The additional authenticated data AAD where 0 <= AAD < 2^64. This additional data is authenticated but not encrypted, and is not included in the output of this mode. It can be used to authenticate plaintext packet headers, or contextual information that affects the interpretation of the message. Users who do not wish to authenticate additional data can provide a string of length zero.

The inputs are summarized as: 

      Name  Description                          Size
      ----  -----------------------------------  -----------------------
      K     Block cipher key                     Depends on block cipher
      N     Nonce                                15-L octets
      m     Message to authenticate and encrypt  l(m) octets
      AAD   Additional authenticated data        0 =< AAD  < 2^64

 TOC 

3.2.4.  Authentication

The first step is to compute the authentication field T. This is done using CBC-MAC [13] (National Institute of Standards and Technology, “Computer Data Authentication,” May 1985.). We first define a sequence of blocks B_0, B_1, ..., B_n and then apply CBC-MAC to these blocks.

The first block B_0 is formatted as follows, where l(m) is encoded in most-significant-byte first order: 

      Octet Number   Contents
      ------------   ---------
      0              Flags
      1 ... 15-L     Nonce N
      16-L ... 15    l(m)

Within the first block B_0, the Flags field is formatted as follows: 

      Bit Number   Contents
      ----------   ----------------------
      7            Reserved (always zero)
      6            Adata
      5 ... 3      M'
      2 ... 0      L'

Another way say the same thing is: Flags = 64*Adata + 8*M' + L'.

The Reserved bit is reserved for future expansions and should always be set to zero. The Adata bit is set to zero if AAD = 0, and set to one if AAD > 0. The M' field is set to (M-2)/2. As M can take on the even values from 4 to 16, the 3-bit M' field can take on the values from one to seven. The 3-bit field MUST NOT have a value of zero, which would correspond to a 16-bit integrity check value. The L' field encodes the size of the length field used to store l(m). The parameter L can take on the values from 2 to 8 (recall, the value L=1 is reserved). This value is encoded in the 3-bit L' field using the values from one to seven by choosing L' = L-1 (the zero value is reserved).

If AAD > 0 (as indicated by the Adata field), then one or more blocks of authentication data are added. These blocks contain AAD and a encoded in a reversible manner. We first construct a string that encodes AAD.

If 0 < AAD < (2^16 - 2^8), then the length field is encoded as two octets which contain the value AAD in most-significant-byte first order.

If (2^16 - 2^8) <= AAD < 2^32, then the length field is encoded as six octets consisting of the octets 0xff, 0xfe, and four octets encoding AAD in most-significant-byte-first order.

If 2^32 <= AAD < 2^64, then the length field is encoded as ten octets consisting of the octets 0xff, 0xff, and eight octets encoding AAD in most-significant-byte-first order.

The length encoding conventions are summarized in the following table. Note that all fields are interpreted in most-significant-byte first order. 

    First two octets   Followed by       Comment
    -----------------  ----------------  -------------------------------
    0x0000             Nothing           Reserved
    0x0001 ... 0xFEFF  Nothing           For 0 < AAD < (2^16 - 2^8)
    0xFF00 ... 0xFFFD  Nothing           Reserved
    0xFFFE             4 octets of AAD   For (2^16 - 2^8) <= AAD < 2^32
    0xFFFF             8 octets of AAD   For 2^32 <= AAD < 2^64

The blocks encoding a are formed by concatenating this string that encodes AAD with a itself, and splitting the result into 16-octet blocks, and then padding the last block with zeroes if necessary. These blocks are appended to the first block B0.

After the (optional) additional authentication blocks have been added, we add the message blocks. The message blocks are formed by splitting the message m into 16-octet blocks, and then padding the last block with zeroes if necessary. If the message m consists of the empty string, then no blocks are added in this step.

The result is a sequence of blocks B0, B1, ..., Bn. The CBC-MAC is computed by: 

      X_1 := Camellia( K, B_0 )
      X_i+1 := Camellia( K, X_i XOR B_i )  for i=1, ..., n
      T := first-M-bytes( X_n+1 )

where T is the MAC value. Note that the last block B_n is XORed with X_n, and the result is encrypted with the block cipher. If needed, the ciphertext is truncated to give T.


 TOC 

3.2.5.  Encryption

To encrypt the message data we use Counter (CTR) mode. We first define the key stream blocks by: 

      S_i := Camellia( K, A_i )   for i=0, 1, 2, ...

The values A_i are formatted as follows, where the Counter field i is encoded in most-significant-byte first order: 

      Octet Number   Contents
      ------------   ---------
      0              Flags
      1 ... 15-L     Nonce N
      16-L ... 15    Counter i

The Flags field is formatted as follows: 

      Bit Number   Contents
      ----------   ----------------------
      7            Reserved (always zero)
      6            Reserved (always zero)
      5 ... 3      Zero
      2 ... 0      L'

Another way say the same thing is: Flags = L'.

The Reserved bits are reserved for future expansions and MUST be set to zero. Bit 6 corresponds to the Adata bit in the B_0 block, but as this bit is not used here, it is reserved and MUST be set to zero. Bits 3, 4, and 5 are also set to zero, ensuring that all the A blocks are distinct from B_0, which has the non-zero encoding of M in this position. Bits 0, 1, and 2 contain L', using the same encoding as in B_0.

The message is encrypted by XORing the octets of message m with the first l(m) octets of the concatenation of S_1, S_2, S_3, ... . Note that S_0 is not used to encrypt the message.

The authentication value U is computed by encrypting T with the key stream block S_0 and truncating it to the desired length. 

      U := T XOR first-M-bytes( S_0 )

 TOC 

3.2.6.  Output

The final result c consists of the encrypted message followed by the encrypted authentication value U.


 TOC 

3.2.7.  Decryption and Authentication Checking

To decrypt a message the following information is required:

  1. The encryption key K.
  2. The nonce N.
  3. The additional authenticated data AAD.
  4. The encrypted and authenticated message c.

Decryption starts by recomputing the key stream to recover the message m and the MAC value T. The message and additional authentication data is then used to recompute the CBC-MAC value and check T.

If the T value is not correct, the receiver MUST NOT reveal any information except for the fact that T is incorrect. The receiver MUST NOT reveal the decrypted message, the value T, or any other information.


 TOC 

4.  Test Vectors


 TOC 

4.1.  Camellia-CTR

This section contains nine test vectors(TV), which can be used to confirm that an implementation has correctly implemented Camellia-CTR. The first three test vectors use Camellia with a 128 bit key; the next three test vectors use Camellia with a 192 bit key; and the last three test vectors use Camellia with a 256 bit key. 


TV #1: Encrypting 16 octets using Camellia-CTR with 128-bit key
Camellia Key     : AE 68 52 F8 12 10 67 CC 4B F7 A5 76 55 77 F3 9E
Camellia-CTR IV  : 00 00 00 00 00 00 00 00
Nonce            : 00 00 00 30
Plaintext        : 53 69 6E 67 6C 65 20 62 6C 6F 63 6B 20 6D 73 67
Counter Block (1): 00 00 00 30 00 00 00 00 00 00 00 00 00 00 00 01
Key Stream    (1): 83 F4 AC FD EE 71 41 F8 4C E8 1F 1D FB 72 78 58
Ciphertext       : D0 9D C2 9A 82 14 61 9A 20 87 7C 76 DB 1F 0B 3F

TV #2: Encrypting 32 octets using Camellia-CTR with 128-bit key
Camellia Key     : 7E 24 06 78 17 FA E0 D7 43 D6 CE 1F 32 53 91 63
Camellia-CTR IV  : C0 54 3B 59 DA 48 D9 0B
Nonce            : 00 6C B6 DB
Plaintext        : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
                 : 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
Counter Block (1): 00 6C B6 DB C0 54 3B 59 DA 48 D9 0B 00 00 00 01
Key Stream    (1): DB F2 C5 8E C4 86 90 D3 D2 75 9A 7C 69 B6 C5 4B
Counter Block (2): 00 6C B6 DB C0 54 3B 59 DA 48 D9 0B 00 00 00 02
Key Stream    (2): 3B 9F 9C 1C 25 E5 CA B0 34 6D 0D F8 4F 7D FE 57
Ciphertext       : DB F3 C7 8D C0 83 96 D4 DA 7C 90 77 65 BB CB 44
                 : 2B 8E 8E 0F 31 F0 DC A7 2C 74 17 E3 53 60 E0 48

TV #3: Encrypting 36 octets using Camellia-CTR with 128-bit key
Camellia Key     : 76 91 BE 03 5E 50 20 A8 AC 6E 61 85 29 F9 A0 DC
Camellia-CTR IV  : 27 77 7F 3F 4A 17 86 F0
Nonce            : 00 E0 01 7B
Plaintext        : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
                 : 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
                 : 20 21 22 23
Counter Block (1): 00 E0 01 7B 27 77 7F 3F 4A 17 86 F0 00 00 00 01
Key Stream    (1): B1 9C 1D CE CF 70 ED 8F 27 8D 96 E9 41 88 C1 7C
Counter Block (2): 00 E0 01 7B 27 77 7F 3F 4A 17 86 F0 00 00 00 02
Key Stream    (2): 8C F7 59 38 48 88 65 E6 57 34 47 86 D2 85 97 D2
Counter Block (3): 00 E0 01 7B 27 77 7F 3F 4A 17 86 F0 00 00 00 03
Key Stream    (3): FF 71 A4 B5 D8 86 12 53 6A 9D 10 A1 13 0F 14 F8
Ciphertext       : B1 9D 1F CD CB 75 EB 88 2F 84 9C E2 4D 85 CF 73
                 : 9C E6 4B 2B 5C 9D 73 F1 4F 2D 5D 9D CE 98 89 CD
                 : DF 50 86 96

TV #4: Encrypting 16 octets using Camellia-CTR with 192-bit key
Camellia Key     : 16 AF 5B 14 5F C9 F5 79 C1 75 F9 3E 3B FB 0E ED
                 : 86 3D 06 CC FD B7 85 15
Camellia-CTR IV  : 36 73 3C 14 7D 6D 93 CB
Nonce            : 00 00 00 48
Plaintext        : 53 69 6E 67 6C 65 20 62 6C 6F 63 6B 20 6D 73 67
Counter Block (1): 00 00 00 48 36 73 3C 14 7D 6D 93 CB 00 00 00 01
Key Stream    (1): 70 10 57 F9 E6 E8 0B 49 7A 1F 4C AC AB F3 E5 F1
Ciphertext       : 23 79 39 9E 8A 8D 2B 2B 16 70 2F C7 8B 9E 96 96

TV #5: Encrypting 32 octets using Camellia-CTR with 192-bit key
Camellia Key     : 7C 5C B2 40 1B 3D C3 3C 19 E7 34 08 19 E0 F6 9C
                 : 67 8C 3D B8 E6 F6 A9 1A
Camellia-CTR IV  : 02 0C 6E AD C2 CB 50 0D
Nonce            : 00 96 B0 3B
Plaintext        : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
                 : 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
Counter Block (1): 00 96 B0 3B 02 0C 6E AD C2 CB 50 0D 00 00 00 01
Key Stream    (1): 7D EE 36 F4 A1 D5 E2 12 6F 42 75 F7 A2 6A C9 52
Counter Block (2): 00 96 B0 3B 02 0C 6E AD C2 CB 50 0D 00 00 00 02
Key Stream    (2): C0 09 AA 7C E6 25 47 F7 4E 20 30 82 EF 47 52 F2
Ciphertext       : 7D EF 34 F7 A5 D0 E4 15 67 4B 7F FC AE 67 C7 5D
                 : D0 18 B8 6F F2 30 51 E0 56 39 2A 99 F3 5A 4C ED

TV #6: Encrypting 36 octets using Camellia-CTR with 192-bit key
Camellia Key     : 02 BF 39 1E E8 EC B1 59 B9 59 61 7B 09 65 27 9B
                 : F5 9B 60 A7 86 D3 E0 FE
Camellia-CTR IV  : 5C BD 60 27 8D CC 09 12
Nonce            : 00 07 BD FD
Plaintext        : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
                 : 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
                 : 20 21 22 23
Counter Block (1): 00 07 BD FD 5C BD 60 27 8D CC 09 12 00 00 00 01
Key Stream    (1): 57 11 E7 55 E5 4D 7C 27 BD A5 04 78 FD 93 40 77
Counter Block (2): 00 07 BD FD 5C BD 60 27 8D CC 09 12 00 00 00 02
Key Stream    (2): 66 E2 6D CF 85 A4 F9 5A 55 B4 F2 FD 7A BB 53 11
Counter Block (3): 00 07 BD FD 5C BD 60 27 8D CC 09 12 00 00 00 03
Key Stream    (3): F5 76 89 74 63 52 A8 C5 1E 82 DE 66 C3 9F 38 34
Ciphertext       : 57 10 E5 56 E1 48 7A 20 B5 AC 0E 73 F1 9E 4E 78
                 : 76 F3 7F DC 91 B1 EF 4D 4D AD E8 E6 66 A6 4D 0E
                 : D5 57 AB 57

TV #7: Encrypting 16 octets using Camellia-CTR with 256-bit key
Camellia Key     : 77 6B EF F2 85 1D B0 6F 4C 8A 05 42 C8 69 6F 6C
                 : 6A 81 AF 1E EC 96 B4 D3 7F C1 D6 89 E6 C1 C1 04
Camellia-CTR IV  : DB 56 72 C9 7A A8 F0 B2
Nonce            : 00 00 00 60
Plaintext        : 53 69 6E 67 6C 65 20 62 6C 6F 63 6B 20 6D 73 67
Counter Block (1): 00 00 00 60 DB 56 72 C9 7A A8 F0 B2 00 00 00 01
Key Stream    (1): 67 68 97 AF 48 1B DF AC D1 06 F7 1A 6C 76 C8 76
Ciphertext       : 34 01 F9 C8 24 7E FF CE BD 69 94 71 4C 1B BB 11

TV #8: Encrypting 32 octets using Camellia-CTR with 256-bit key
Camellia Key     : F6 D6 6D 6B D5 2D 59 BB 07 96 36 58 79 EF F8 86
                 : C6 6D D5 1A 5B 6A 99 74 4B 50 59 0C 87 A2 38 84
Camellia-CTR IV  : C1 58 5E F1 5A 43 D8 75
Nonce            : 00 FA AC 24
Plaintext        : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
                 : 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
Counter Block (1): 00 FA AC 24 C1 58 5E F1 5A 43 D8 75 00 00 00 01
Key Stream    (1): D6 C2 01 91 20 6A 7E 0F A0 35 21 29 A4 8E 90 4A
Counter Block (2): 00 FA AC 24 C1 58 5E F1 5A 43 D8 75 00 00 00 02
Key Stream    (2): F5 0D C6 99 08 CA 56 79 A4 85 D8 C8 B7 9E 5F 17
Ciphertext       : D6 C3 03 92 24 6F 78 08 A8 3C 2B 22 A8 83 9E 45
                 : E5 1C D4 8A 1C DF 40 6E BC 9C C2 D3 AB 83 41 08

TV #9: Encrypting 36 octets using Camellia-CTR with 256-bit key
Camellia Key     : FF 7A 61 7C E6 91 48 E4 F1 72 6E 2F 43 58 1D E2
                 : AA 62 D9 F8 05 53 2E DF F1 EE D6 87 FB 54 15 3D
Camellia-CTR IV  : 51 A5 1D 70 A1 C1 11 48
Nonce            : 00 1C C5 B7
Plaintext        : 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
                 : 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
                 : 20 21 22 23
Counter Block (1): 00 1C C5 B7 51 A5 1D 70 A1 C1 11 48 00 00 00 01
Key Stream    (1): A4 DB 21 FF E2 A0 F9 AD 65 6D A4 91 0A 5F AA 23
Counter Block (2): 00 1C C5 B7 51 A5 1D 70 A1 C1 11 48 00 00 00 02
Key Stream    (2): C1 70 B1 58 71 EC 71 88 6D D9 05 0B 03 6C 39 70
Counter Block (3): 00 1C C5 B7 51 A5 1D 70 A1 C1 11 48 00 00 00 03
Key Stream    (3): 35 CE 2F AE 90 78 B3 72 F5 76 12 39 1F 8B AF BF
Ciphertext       : A4 DA 23 FC E6 A5 FF AA 6D 64 AE 9A 06 52 A4 2C
                 : D1 61 A3 4B 65 F9 67 9F 75 C0 1F 10 1F 71 27 6F
                 : 15 EF 0D 8D

 TOC 

4.2.  Camellia-CCM

This section contains twenty four test vectors, which can be used to confirm that an implementation has correctly implemented Camellia-CCM. In each of these test vectors, the least significant sixteen bits of the counter block is used for the block counter, and the nonce is 13 octets. Some of the test vectors include a eight octet authentication value, and others include a ten octet authentication value. 

=============== Packet Vector #1 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 03  02 01 00 A0  A1 A2 A3 A4  A5
Total packet length =   31. [Input (8 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E
CBC IV in: 59 00 00 00  03 02 01 00  A0 A1 A2 A3  A4 A5 00 17
CBC IV out:D4 DB CD 92  A8 96 41 56  1D 0D BB D0  D5 7F 7E 1D
After xor: D4 D3 CD 93  AA 95 45 53  1B 0A BB D0  D5 7F 7E 1D   [hdr]
After CAM: BD 84 03 80  73 59 37 B7  CE F5 E4 BA  1B 18 54 DC
After xor: B5 8D 09 8B  7F 54 39 B8  DE E4 F6 A9  0F 0D 42 CB   [msg]
After CAM: CE 21 82 9C  F6 F2 4D A2  CB 35 D1 FD  81 27 63 EC
After xor: D6 38 98 87  EA EF 53 A2  CB 35 D1 FD  81 27 63 EC   [msg]
After CAM: 20 11 FE E2  53 B1 A7 DB  02 77 FA 37  6D 78 EE 10
MIC tag  : 20 11 FE E2  53 B1 A7 DB
CTR Start: 01 00 00 00  03 02 01 00  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: B2 7A 7B 8E  EB 14 3F 0B  82 E2 98 4C  06 44 CC 42
CTR[0002]: E2 E2 D3 52  98 97 13 45  D1 63 22 90  E7 F8 15 4A
CTR[MIC ]: DC BF 30 96  38 8C 1E 76
Total packet length =   39. [Encrypted]
           00 01 02 03  04 05 06 07  BA 73 71 85  E7 19 31 04
           92 F3 8A 5F  12 51 DA 55  FA FB C9 49  84 8A 0D FC
           AE CE 74 6B  3D B9 AD

=============== Packet Vector #2 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 04  03 02 01 A0  A1 A2 A3 A4  A5
Total packet length =   32. [Input (8 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E 1F
CBC IV in: 59 00 00 00  04 03 02 01  A0 A1 A2 A3  A4 A5 00 18
CBC IV out:07 0B 22 50  8A 24 3C DD  5B BA 54 DB  60 52 88 06
After xor: 07 03 22 51  88 27 38 D8  5D BD 54 DB  60 52 88 06   [hdr]
After CAM: 10 FD C2 F2  90 4A 9F 96  B0 4F 62 A4  A1 A9 31 1E
After xor: 18 F4 C8 F9  9C 47 91 99  A0 5E 70 B7  B5 BC 27 09   [msg]
After CAM: E4 C8 82 02  89 55 5C 15  CE 7F E4 60  B1 B9 5A 08
After xor: FC D1 98 19  95 48 42 0A  CE 7F E4 60  B1 B9 5A 08   [msg]
After CAM: D2 96 BA 4F  83 DE B5 DF  A2 19 08 F7  47 4E 3C 40
MIC tag  : D2 96 BA 4F  83 DE B5 DF
CTR Start: 01 00 00 00  04 03 02 01  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: 55 2C 6E B4  82 A2 EF D6  85 37 FE 12  79 0E E6 55
CTR[0002]: 54 E2 C8 D6  7E 99 91 2C  F2 8A D7 8E  83 04 10 36
CTR[MIC ]: B2 24 93 12  71 9C 36 37
Total packet length =   40. [Encrypted]
           00 01 02 03  04 05 06 07  5D 25 64 BF  8E AF E1 D9
           95 26 EC 01  6D 1B F0 42  4C FB D2 CD  62 84 8F 33
           60 B2 29 5D  F2 42 83 E8

=============== Packet Vector #3 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 05  04 03 02 A0  A1 A2 A3 A4  A5
Total packet length =   33. [Input (8 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E 1F
           20
CBC IV in: 59 00 00 00  05 04 03 02  A0 A1 A2 A3  A4 A5 00 19
CBC IV out:6F 69 15 DF  A6 A0 DF 24  84 A7 37 88  A3 65 F9 2E
After xor: 6F 61 15 DE  A4 A3 DB 21  82 A0 37 88  A3 65 F9 2E   [hdr]
After CAM: 59 5D 99 48  79 04 DA C9  13 93 36 C9  11 A8 09 1D
After xor: 51 54 93 43  75 09 D4 C6  03 82 24 DA  05 BD 1F 0A   [msg]
After CAM: 1A 43 D7 19  65 43 97 C1  43 6F 4F 11  A7 6C 6B ED
After xor: 02 5A CD 02  79 5E 89 DE  63 6F 4F 11  A7 6C 6B ED   [msg]
After CAM: 30 0B 06 8A  A0 D1 4D C5  9E 44 22 84  82 45 42 0B
MIC tag  : 30 0B 06 8A  A0 D1 4D C5
CTR Start: 01 00 00 00  05 04 03 02  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: 89 FF 69 DD  CB 75 76 18  E9 31 24 1B  AD 97 BB 02
CTR[0002]: C4 32 A7 9C  CB 4B E9 8D  24 A8 F0 AB  C6 87 16 11
CTR[MIC ]: C5 5A D0 E2  8F F2 E7 83
Total packet length =   41. [Encrypted]
           00 01 02 03  04 05 06 07  81 F6 63 D6  C7 78 78 17
           F9 20 36 08  B9 82 AD 15  DC 2B BD 87  D7 56 F7 92
           04 F5 51 D6  68 2F 23 AA  46

=============== Packet Vector #4 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 06  05 04 03 A0  A1 A2 A3 A4  A5
Total packet length =   31. [Input (12 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E
CBC IV in: 59 00 00 00  06 05 04 03  A0 A1 A2 A3  A4 A5 00 13
CBC IV out:F5 51 CF 6C  7C F7 D4 0B  2B 76 F1 6B  57 F0 19 FE
After xor: F5 5D CF 6D  7E F4 D0 0E  2D 71 F9 62  5D FB 19 FE   [hdr]
After CAM: 02 2B 21 1B  EB 97 02 3B  F8 10 7D CC  62 14 E5 7C
After xor: 0E 26 2F 14  FB 86 10 28  EC 05 6B DB  7A 0D FF 67   [msg]
After CAM: 48 14 A4 2D  31 25 1C 37  19 C5 6F DD  5A 37 81 42
After xor: 54 09 BA 2D  31 25 1C 37  19 C5 6F DD  5A 37 81 42   [msg]
After CAM: CF 85 25 D2  80 D5 F0 09  53 2C 9D 43  4E F3 04 47
MIC tag  : CF 85 25 D2  80 D5 F0 09
CTR Start: 01 00 00 00  06 05 04 03  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: C6 E2 10 8D  62 00 A2 9C  6F CC 19 1F  DF 6B 92 DB
CTR[0002]: 6C B9 BE EE  1E A2 E9 B3  2D D6 C2 9A  E8 26 D5 C2
CTR[MIC ]: 44 BF B6 E8  E3 31 67 A9
Total packet length =   39. [Encrypted]
           00 01 02 03  04 05 06 07  08 09 0A 0B  CA EF 1E 82
           72 11 B0 8F  7B D9 0F 08  C7 72 88 C0  70 A4 A0 8B
           3A 93 3A 63  E4 97 A0

=============== Packet Vector #5 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 07  06 05 04 A0  A1 A2 A3 A4  A5
Total packet length =   32. [Input (12 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E 1F
CBC IV in: 59 00 00 00  07 06 05 04  A0 A1 A2 A3  A4 A5 00 14
CBC IV out:73 72 9D 76  7A BD B9 82  60 3A 12 7B  EF 26 FB 80
After xor: 73 7E 9D 77  78 BE BD 87  66 3D 1A 72  E5 2D FB 80   [hdr]
After CAM: E1 B7 A6 72  E2 5C 87 75  91 21 22 A4  07 13 CD 5B
After xor: ED BA A8 7D  F2 4D 95 66  85 34 34 B3  1F 0A D7 40   [msg]
After CAM: 13 2F 58 D9  5D 0F 95 B8  90 BF 6F 1D  31 84 54 C7
After xor: 0F 32 46 C6  5D 0F 95 B8  90 BF 6F 1D  31 84 54 C7   [msg]
After CAM: 47 8F 1E B0  71 24 8B 13  AF C8 C8 44  E6 0F 88 B6
MIC tag  : 47 8F 1E B0  71 24 8B 13
CTR Start: 01 00 00 00  07 06 05 04  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: 26 DE B4 D6  5F D4 3C 81  AA 56 98 95  64 09 39 A2
CTR[0002]: 76 97 69 3A  21 13 0C 39  2E 4E EB BF  48 7B 24 BE
CTR[MIC ]: C8 2E 65 17  82 15 50 1A
Total packet length =   40. [Encrypted]
           00 01 02 03  04 05 06 07  08 09 0A 0B  2A D3 BA D9
           4F C5 2E 92  BE 43 8E 82  7C 10 23 B9  6A 8A 77 25
           8F A1 7B A7  F3 31 DB 09

=============== Packet Vector #6 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 08  07 06 05 A0  A1 A2 A3 A4  A5
Total packet length =   33. [Input (12 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E 1F
           20
CBC IV in: 59 00 00 00  08 07 06 05  A0 A1 A2 A3  A4 A5 00 15
CBC IV out:EB 59 05 CC  3F 52 61 10  26 24 75 93  DD B9 A0 F4
After xor: EB 55 05 CD  3D 51 65 15  20 23 7D 9A  D7 B2 A0 F4   [hdr]
After CAM: 18 A9 AE A4  3D D2 A9 11  6C 0A E5 4F  40 D1 4D 9F
After xor: 14 A4 A0 AB  2D C3 BB 02  78 1F F3 58  58 C8 57 84   [msg]
After CAM: FA C4 13 18  98 54 1B 54  93 9C 64 B8  CB FD 5B 18
After xor: E6 D9 0D 07  B8 54 1B 54  93 9C 64 B8  CB FD 5B 18   [msg]
After CAM: 49 E6 E8 ED  32 FB CA 2F  2E 55 CD AF  D0 F2 B3 05
MIC tag  : 49 E6 E8 ED  32 FB CA 2F
CTR Start: 01 00 00 00  08 07 06 05  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: F2 A8 46 04  B5 2E BA C0  D7 51 34 BD  D6 54 FC 64
CTR[0002]: E6 26 A9 24  8B E6 86 CB  92 D6 FB FC  2E F2 91 98
CTR[MIC ]: E2 D0 49 03  7D 1B 34 07
Total packet length =   41. [Encrypted]
           00 01 02 03  04 05 06 07  08 09 0A 0B  FE A5 48 0B
           A5 3F A8 D3  C3 44 22 AA  CE 4D E6 7F  FA 3B B7 3B
           AB AB 36 A1  EE 4F E0 FE  28

=============== Packet Vector #7 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 09  08 07 06 A0  A1 A2 A3 A4  A5
Total packet length =   31. [Input (8 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E
CBC IV in: 61 00 00 00  09 08 07 06  A0 A1 A2 A3  A4 A5 00 17
CBC IV out:AC F1 5D 79  99 1A 15 BF  5C DC F6 C4  45 AE 1F CB
After xor: AC F9 5D 78  9B 19 11 BA  5A DB F6 C4  45 AE 1F CB   [hdr]
After CAM: E9 C0 AC FD  C7 E8 E7 1D  FA E8 8B 66  95 9E 01 45
After xor: E1 C9 A6 F6  CB E5 E9 12  EA F9 99 75  81 8B 17 52   [msg]
After CAM: 9C FF ED 72  09 A6 7D 2A  48 B7 29 BF  D8 BE 39 59
After xor: 84 E6 F7 69  15 BB 63 2A  48 B7 29 BF  D8 BE 39 59   [msg]
After CAM: 4F 41 FA DE  B2 58 F3 32  54 0A 55 7A  80 4A A3 F5
MIC tag  : 4F 41 FA DE  B2 58 F3 32  54 0A
CTR Start: 01 00 00 00  09 08 07 06  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: 5C 5A 2A 2D  E9 41 1F 95  9D 27 CB FF  7A 0B CF 63
CTR[0002]: 0E D1 6A 97  57 41 32 4F  33 1B 4A 42  B1 4A 54 63
CTR[MIC ]: E3 EE 59 62  7D 22 BD 8D  C1 79
Total packet length =   41. [Encrypted]
           00 01 02 03  04 05 06 07  54 53 20 26  E5 4C 11 9A
           8D 36 D9 EC  6E 1E D9 74  16 C8 70 8C  4B 5C 2C AC
           AF A3 BC CF  7A 4E BF 95  73

=============== Packet Vector #8 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 0A  09 08 07 A0  A1 A2 A3 A4  A5
Total packet length =   32. [Input (8 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E 1F
CBC IV in: 61 00 00 00  0A 09 08 07  A0 A1 A2 A3  A4 A5 00 18
CBC IV out:AD CA 1C 1D  45 E7 E2 62  58 D5 DA 46  D8 2F 69 3A
After xor: AD C2 1C 1C  47 E4 E6 67  5E D2 DA 46  D8 2F 69 3A   [hdr]
After CAM: FA DE 0E B4  3E CA C1 E9  69 BB 8C A4  7C 0D 80 8F
After xor: F2 D7 04 BF  32 C7 CF E6  79 AA 9E B7  68 18 96 98   [msg]
After CAM: D2 87 35 C2  D0 E4 AE 4E  BC C2 99 FF  B3 77 F8 A1
After xor: CA 9E 2F D9  CC F9 B0 51  BC C2 99 FF  B3 77 F8 A1   [msg]
After CAM: BD F6 FB 55  9E 90 C0 E7  DF 4B 0C 37  DC 42 32 A2
MIC tag  : BD F6 FB 55  9E 90 C0 E7  DF 4B
CTR Start: 01 00 00 00  0A 09 08 07  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: 82 D8 91 0B  16 8A DF 47  E4 C8 39 FC  20 47 4A DB
CTR[0002]: FB BF 26 7E  0E BB EB 6A  07 4E 29 CF  3D 12 E6 DB
CTR[MIC ]: CE 7E 1F C4  A0 61 87 E6  2B 0A
Total packet length =   42. [Encrypted]
           00 01 02 03  04 05 06 07  8A D1 9B 00  1A 87 D1 48
           F4 D9 2B EF  34 52 5C CC  E3 A6 3C 65  12 A6 F5 75
           73 88 E4 91  3E F1 47 01  F4 41

=============== Packet Vector #9 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 0B  0A 09 08 A0  A1 A2 A3 A4  A5
Total packet length =   33. [Input (8 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E 1F
           20
CBC IV in: 61 00 00 00  0B 0A 09 08  A0 A1 A2 A3  A4 A5 00 19
CBC IV out:D0 A9 A5 94  00 63 86 40  11 0D DB 40  CA F8 4A 9C
After xor: D0 A1 A5 95  02 60 82 45  17 0A DB 40  CA F8 4A 9C   [hdr]
After CAM: 7B CA 4E 2D  79 82 0D 1E  15 22 DD E8  37 B9 B1 F0
After xor: 73 C3 44 26  75 8F 03 11  05 33 CF FB  23 AC A7 E7   [msg]
After CAM: 6B 75 9F 83  C0 8F 56 64  F2 FA D5 7F  67 01 B8 21
After xor: 73 6C 85 98  DC 92 48 7B  D2 FA D5 7F  67 01 B8 21   [msg]
After CAM: 7D B7 BE FF  72 F3 26 74  9E 20 07 28  1E 5B 1A 8A
MIC tag  : 7D B7 BE FF  72 F3 26 74  9E 20
CTR Start: 01 00 00 00  0B 0A 09 08  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: 55 B9 87 69  4C 73 60 3E  C6 1E 8E B1  D2 11 62 36
CTR[0002]: 82 D9 A4 4B  DC C9 BB 68  A7 FE 15 A5  19 51 57 87
CTR[MIC ]: E9 61 5C CF  BF D6 EF 8A  21 A7
Total packet length =   43. [Encrypted]
           00 01 02 03  04 05 06 07  5D B0 8D 62  40 7E 6E 31
           D6 0F 9C A2  C6 04 74 21  9A C0 BE 50  C0 D4 A5 77
           87 94 D6 E2  30 CD 25 C9  FE BF 87

=============== Packet Vector #10 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 0C  0B 0A 09 A0  A1 A2 A3 A4  A5
Total packet length =   31. [Input (12 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E
CBC IV in: 61 00 00 00  0C 0B 0A 09  A0 A1 A2 A3  A4 A5 00 13
CBC IV out:B1 85 73 A3  1C 6F EC 01  90 E3 CE 94  27 11 04 B9
After xor: B1 89 73 A2  1E 6C E8 04  96 E4 C6 9D  2D 1A 04 B9   [hdr]
After CAM: A6 AD EA 9C  FA 3F 76 78  4C 17 8A F3  DC 69 F0 82
After xor: AA A0 E4 93  EA 2E 64 6B  58 02 9C E4  C4 70 EA 99   [msg]
After CAM: 35 50 B7 27  78 F8 C6 BF  02 4B 65 60  05 C0 E1 ED
After xor: 29 4D A9 27  78 F8 C6 BF  02 4B 65 60  05 C0 E1 ED   [msg]
After CAM: 3D B5 A6 E6  85 AF 1C 58  80 B0 32 2E  01 74 91 FC
MIC tag  : 3D B5 A6 E6  85 AF 1C 58  80 B0
CTR Start: 01 00 00 00  0C 0B 0A 09  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: D7 1C 82 C1  D1 A9 64 0F  93 69 CE 81  22 7E CC E8
CTR[0002]: A7 A1 42 44  32 4E 69 FE  4C D0 36 65  A5 31 0B AB
CTR[MIC ]: ED 27 3F 0D  94 5C 0E AA  B2 87
Total packet length =   41. [Encrypted]
           00 01 02 03  04 05 06 07  08 09 0A 0B  DB 11 8C CE
           C1 B8 76 1C  87 7C D8 96  3A 67 D6 F3  BB BC 5C D0
           92 99 EB 11  F3 12 F2 32  37

=============== Packet Vector #11 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 0D  0C 0B 0A A0  A1 A2 A3 A4  A5
Total packet length =   32. [Input (12 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E 1F
CBC IV in: 61 00 00 00  0D 0C 0B 0A  A0 A1 A2 A3  A4 A5 00 14
CBC IV out:45 DF B5 07  6F BB 10 EA  F1 15 15 AD  21 4F B0 0E
After xor: 45 D3 B5 06  6D B8 14 EF  F7 12 1D A4  2B 44 B0 0E   [hdr]
After CAM: 17 52 F9 6D  DD BC 5B 1C  1E EB 80 FC  F6 10 AC 03
After xor: 1B 5F F7 62  CD AD 49 0F  0A FE 96 EB  EE 09 B6 18   [msg]
After CAM: BE F0 A0 B9  EC 94 B6 B3  E8 EC 1B 82  14 14 09 87
After xor: A2 ED BE A6  EC 94 B6 B3  E8 EC 1B 82  14 14 09 87   [msg]
After CAM: 70 16 E4 F9  C4 2C 30 10  84 BF EC 69  34 89 91 FD
MIC tag  : 70 16 E4 F9  C4 2C 30 10  84 BF
CTR Start: 01 00 00 00  0D 0C 0B 0A  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: 70 C5 33 82  D4 80 11 41  4F 5D 2B D2  D2 67 B3 B0
CTR[0002]: 9D 36 6E 49  39 C5 16 76  5C 1C 25 12  81 79 94 70
CTR[MIC ]: 77 8B 4B 03  1E 3A FC DF  A8 F1
Total packet length =   42. [Encrypted]
           00 01 02 03  04 05 06 07  08 09 0A 0B  7C C8 3D 8D
           C4 91 03 52  5B 48 3D C5  CA 7E A9 AB  81 2B 70 56
           07 9D AF FA  DA 16 CC CF  2C 4E

=============== Packet Vector #12 ==================
CAM Key:   C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 0E  0D 0C 0B A0  A1 A2 A3 A4  A5
Total packet length =   33. [Input (12 cleartext header octets)]
           00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
           10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E 1F
           20
CBC IV in: 61 00 00 00  0E 0D 0C 0B  A0 A1 A2 A3  A4 A5 00 15
CBC IV out:81 E4 EB 1E  50 A9 70 CE  18 CA 1A 4B  68 39 80 2E
After xor: 81 E8 EB 1F  52 AA 74 CB  1E CD 12 42  62 32 80 2E   [hdr]
After CAM: 04 AB D9 62  34 B9 8F 32  8C 0F 08 3F  3D 87 9D 57
After xor: 08 A6 D7 6D  24 A8 9D 21  98 1A 1E 28  25 9E 87 4C   [msg]
After CAM: BD A2 EA CB  3A DA 6A E7  9F BB C2 2C  E6 4C 98 89
After xor: A1 BF F4 D4  1A DA 6A E7  9F BB C2 2C  E6 4C 98 89   [msg]
After CAM: B6 FC E1 46  D3 EA DC 91  E0 AB 10 AD  D8 55 E7 03
MIC tag  : B6 FC E1 46  D3 EA DC 91  E0 AB
CTR Start: 01 00 00 00  0E 0D 0C 0B  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: 20 DE 55 87  30 C3 2C 69  B7 44 A6 FE  37 DE 89 7C
CTR[0002]: 3F 96 32 D8  68 6D C2 B5  22 97 42 27  EB F9 26 5E
CTR[MIC ]: 7D 45 AD 6F  94 93 E1 F5  4F DE
Total packet length =   43. [Encrypted]
           00 01 02 03  04 05 06 07  08 09 0A 0B  2C D3 5B 88
           20 D2 3E 7A  A3 51 B0 E9  2F C7 93 67  23 8B 2C C7
           48 CB B9 4C  29 47 79 3D  64 AF 75

=============== Packet Vector #13 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 A9 70 11  0E 19 27 B1  60 B6 A3 1C  1C
Total packet length =   31. [Input (8 cleartext header octets)]
           6B 7F 46 45  07 FA E4 96  C6 B5 F3 E6  CA 23 11 AE
           F7 47 2B 20  3E 73 5E A5  61 AD B1 7D  56 C5 A3
CBC IV in: 59 00 A9 70  11 0E 19 27  B1 60 B6 A3  1C 1C 00 17
CBC IV out:D7 24 B0 0F  B1 87 04 C6  C1 4E 90 37  AA F2 F1 F9
After xor: D7 2C DB 70  F7 C2 03 3C  25 D8 90 37  AA F2 F1 F9   [hdr]
After CAM: 9B 13 6D E3  D9 9F C3 6D  7D 0D B7 D8  A1 BF E9 BD
After xor: 5D A6 9E 05  13 BC D2 C3  8A 4A 9C F8  9F CC B7 18   [msg]
After CAM: F8 BF 25 7D  23 F8 D9 B5  82 E6 C9 3E  C8 9B 85 73
After xor: 99 12 94 00  75 3D 7A B5  82 E6 C9 3E  C8 9B 85 73   [msg]
After CAM: D9 D6 62 21  6D B2 CA FD  1F C6 FE 9D  2C AF 5B 69
MIC tag  : D9 D6 62 21  6D B2 CA FD
CTR Start: 01 00 A9 70  11 0E 19 27  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: 62 80 24 C1  FE AE CC 8C  67 38 55 98  CB 8E E5 E8
CTR[0002]: F2 30 17 2F  1B 71 55 9F  8B CE 79 E5  13 01 FC 6A
CTR[MIC ]: 9C 8E A2 0C  48 03 ED 13
Total packet length =   39. [Encrypted]
           6B 7F 46 45  07 FA E4 96  A4 35 D7 27  34 8D DD 22
           90 7F 7E B8  F5 FD BB 4D  93 9D A6 52  4D B4 F6 45
           58 C0 2D 25  B1 27 EE

=============== Packet Vector #14 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 83 CD 8C  E0 CB 42 B1  60 B6 A3 1C  1C
Total packet length =   32. [Input (8 cleartext header octets)]
           98 66 05 B4  3D F1 5D E7  01 F6 CE 67  64 C5 74 48
           3B B0 2E 6B  BF 1E 0A BD  26 A2 25 72  B4 D8 0E E7
CBC IV in: 59 00 83 CD  8C E0 CB 42  B1 60 B6 A3  1C 1C 00 18
CBC IV out:A0 8A 29 78  36 23 1D 84  96 76 93 FF  0A 4C 92 7A
After xor: A0 82 B1 1E  33 97 20 75  CB 91 93 FF  0A 4C 92 7A   [hdr]
After CAM: 8C F5 F4 23  BF 09 1C 74  CD 47 00 C1  32 5D 5C 92
After xor: 8D 03 3A 44  DB CC 68 3C  F6 F7 2E AA  8D 43 56 2F   [msg]
After CAM: 69 DA 48 24  41 1E AC 8E  A9 0A CD 8B  DD 00 2B 9A
After xor: 4F 78 6D 56  F5 C6 A2 69  A9 0A CD 8B  DD 00 2B 9A   [msg]
After CAM: C2 03 3B 08  6D B3 CB 3B  2C C8 5D E7  76 A1 C0 44
MIC tag  : C2 03 3B 08  6D B3 CB 3B
CTR Start: 01 00 83 CD  8C E0 CB 42  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: 8B 16 9C 37  EB 7B BE DB  15 84 41 6E  5F C2 07 46
CTR[0002]: E9 31 BB DD  4E E6 56 9B  68 95 13 5F  AB A4 DF EF
CTR[MIC ]: 44 7E 55 14  25 C3 F3 3D
Total packet length =   40. [Encrypted]
           98 66 05 B4  3D F1 5D E7  8A E0 52 50  8F BE CA 93
           2E 34 6F 05  E0 DC 0D FB  CF 93 9E AF  FA 3E 58 7C
           86 7D 6E 1C  48 70 38 06

=============== Packet Vector #15 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 5F 54 95  0B 18 F2 B1  60 B6 A3 1C  1C
Total packet length =   33. [Input (8 cleartext header octets)]
           48 F2 E7 E1  A7 67 1A 51  CD F1 D8 40  6F C2 E9 01
           49 53 89 70  05 FB FB 8B  A5 72 76 F9  24 04 60 8E
           08
CBC IV in: 59 00 5F 54  95 0B 18 F2  B1 60 B6 A3  1C 1C 00 19
CBC IV out:76 74 53 37  95 23 3C F0  EB 77 CE 93  73 06 99 A8
After xor: 76 7C 1B C5  72 C2 9B 97  F1 26 CE 93  73 06 99 A8   [hdr]
After CAM: EF 79 8B 70  34 E4 D5 6B  57 3A F9 44  F0 AF D6 9A
After xor: 22 88 53 30  5B 26 3C 6A  1E 69 70 34  F5 54 2D 11   [msg]
After CAM: 63 BF 4E 10  01 79 38 0B  E4 EC C1 39  B2 B4 3B 8C
After xor: C6 CD 38 E9  25 7D 58 85  EC EC C1 39  B2 B4 3B 8C   [msg]
After CAM: 39 E1 0E FA  BD 2F 43 00  50 9E E7 EB  A4 FF 6B 8F
MIC tag  : 39 E1 0E FA  BD 2F 43 00
CTR Start: 01 00 5F 54  95 0B 18 F2  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: C5 47 A6 A2  73 49 1B 6F  0E 6D C9 F5  9C 12 3B 08
CTR[0002]: C8 18 86 42  3C DB 35 C8  64 4D 8C 4C  58 01 47 27
CTR[MIC ]: 91 E9 76 5D  2D 68 2E E5
Total packet length =   41. [Encrypted]
           48 F2 E7 E1  A7 67 1A 51  08 B6 7E E2  1C 8B F2 6E
           47 3E 40 85  99 E9 C0 83  6D 6A F0 BB  18 DF 55 46
           6C A8 08 78  A7 90 47 6D  E5

=============== Packet Vector #16 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 EC 60 08  63 31 9A B1  60 B6 A3 1C  1C
Total packet length =   31. [Input (12 cleartext header octets)]
           DE 97 DF 3B  8C BD 6D 8E  50 30 DA 4C  B0 05 DC FA
           0B 59 18 14  26 A9 61 68  5A 99 3D 8C  43 18 5B
CBC IV in: 59 00 EC 60  08 63 31 9A  B1 60 B6 A3  1C 1C 00 13
CBC IV out:78 EE 05 5A  88 48 E3 5B  8A 45 46 8F  35 4F 0C A2
After xor: 78 E2 DB CD  57 73 6F E6  E7 CB 16 BF  EF 03 0C A2   [hdr]
After CAM: A9 C6 7F 15  00 1A C6 92  81 67 BD EC  DF D2 35 C9
After xor: 19 C3 A3 EF  0B 43 DE 86  A7 CE DC 84  85 4B 08 45   [msg]
After CAM: 7C A8 9C 90  46 42 4B E2  4D 96 DF CF  BA 12 FD 18
After xor: 3F B0 C7 90  46 42 4B E2  4D 96 DF CF  BA 12 FD 18   [msg]
After CAM: 89 C7 B4 E8  A4 24 8C 6C  52 ED 34 50  E3 53 AD F5
MIC tag  : 89 C7 B4 E8  A4 24 8C 6C
CTR Start: 01 00 EC 60  08 63 31 9A  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: D3 B2 57 B3  6C E8 86 CF  91 9A AC 79  4E 6F 73 3E
CTR[0002]: 65 10 C8 72  39 AF 0F 52  9F D0 A4 DF  54 BF D6 EB
CTR[MIC ]: E1 04 E0 6A  29 B1 80 A9
Total packet length =   39. [Encrypted]
           DE 97 DF 3B  8C BD 6D 8E  50 30 DA 4C  63 B7 8B 49
           67 B1 9E DB  B7 33 CD 11  14 F6 4E B2  26 08 93 68
           C3 54 82 8D  95 0C C5

=============== Packet Vector #17 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 60 CF F1  A3 1E A1 B1  60 B6 A3 1C  1C
Total packet length =   32. [Input (12 cleartext header octets)]
           A5 EE 93 E4  57 DF 05 46  6E 78 2D CF  2E 20 21 12
           98 10 5F 12  9D 5E D9 5B  93 F7 2D 30  B2 FA CC D7
CBC IV in: 59 00 60 CF  F1 A3 1E A1  B1 60 B6 A3  1C 1C 00 14
CBC IV out:C3 34 69 7D  11 38 73 06  BD 34 E2 10  1F 66 17 E8
After xor: C3 38 CC 93  82 DC 24 D9  B8 72 8C 68  32 A9 17 E8   [hdr]
After CAM: 43 6F 37 74  AB 94 3B 41  EA AD 00 CA  C3 99 13 7B
After xor: 6D 4F 16 66  33 84 64 53  77 F3 D9 91  50 6E 3E 4B   [msg]
After CAM: 2D 28 FB 62  DA 06 97 A7  4C D4 31 B8  B5 AE AE EE
After xor: 9F D2 37 B5  DA 06 97 A7  4C D4 31 B8  B5 AE AE EE   [msg]
After CAM: F3 DE 10 CD  91 4D B1 B6  CC 37 F0 A2  4A 5A B7 A1
MIC tag  : F3 DE 10 CD  91 4D B1 B6
CTR Start: 01 00 60 CF  F1 A3 1E A1  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: 25 E6 9A F0  30 A9 56 E6  FF C0 3F 87  87 7A 89 74
CTR[0002]: A2 1B 46 23  76 A2 1E DD  F2 AC 4B EC  42 95 3D D3
CTR[MIC ]: C2 99 28 FF  E7 BB DB 29
Total packet length =   40. [Encrypted]
           A5 EE 93 E4  57 DF 05 46  6E 78 2D CF  0B C6 BB E2
           A8 B9 09 F4  62 9E E6 DC  14 8D A4 44  10 E1 8A F4
           31 47 38 32  76 F6 6A 9F

=============== Packet Vector #18 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 0F 85 CD  99 5C 97 B1  60 B6 A3 1C  1C
Total packet length =   33. [Input (12 cleartext header octets)]
           24 AA 1B F9  A5 CD 87 61  82 A2 50 74  26 45 94 1E
           75 63 2D 34  91 AF 0F C0  C9 87 6C 3B  E4 AA 74 68
           C9
CBC IV in: 59 00 0F 85  CD 99 5C 97  B1 60 B6 A3  1C 1C 00 15
CBC IV out:72 0A 46 75  0F 40 59 53  F2 3B D2 1F  6A 11 60 F6
After xor: 72 06 62 DF  14 B9 FC 9E  75 5A 50 BD  3A 65 60 F6   [hdr]
After CAM: 67 73 A0 FD  D5 7E D3 5E  E8 24 06 D0  A1 8B 0E 18
After xor: 41 36 34 E3  A0 1D FE 6A  79 8B 09 10  68 0C 62 23   [msg]
After CAM: BB 1E D8 9F  60 29 D0 99  09 14 06 A5  E3 8B 72 7B
After xor: 5F B4 AC F7  A9 29 D0 99  09 14 06 A5  E3 8B 72 7B   [msg]
After CAM: 3E 4F 40 73  D1 31 E9 B8  02 C8 99 BC  FD AC 19 4B
MIC tag  : 3E 4F 40 73  D1 31 E9 B8
CTR Start: 01 00 0F 85  CD 99 5C 97  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: 04 6F 42 2C  8F 52 FB 9B  06 A3 3B 9F  B7 F0 A6 00
CTR[0002]: 34 76 51 DB  89 10 FB E6  73 E8 56 6E  DB 66 47 5D
CTR[MIC ]: 9F EC 93 6C  5C 7A AD 0F
Total packet length =   41. [Encrypted]
           24 AA 1B F9  A5 CD 87 61  82 A2 50 74  22 2A D6 32
           FA 31 D6 AF  97 0C 34 5F  7E 77 CA 3B  D0 DC 25 B3
           40 A1 A3 D3  1F 8D 4B 44  B7

=============== Packet Vector #19 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 C2 9B 2C  AA C4 CD B1  60 B6 A3 1C  1C
Total packet length =   31. [Input (8 cleartext header octets)]
           69 19 46 B9  CA 07 BE 87  07 01 35 A6  43 7C 9D B1
           20 CD 61 D8  F6 C3 9C 3E  A1 25 FD 95  A0 D2 3D
CBC IV in: 61 00 C2 9B  2C AA C4 CD  B1 60 B6 A3  1C 1C 00 17
CBC IV out:74 AD F8 04  05 2A 48 E7  46 97 38 D5  BA A1 27 79
After xor: 74 A5 91 1D  43 93 82 E0  F8 10 38 D5  BA A1 27 79   [hdr]
After CAM: BD C3 B1 41  1C 64 C8 B3  A9 DC 6A 94  78 97 88 E2
After xor: BA C2 84 E7  5F 18 55 02  89 11 0B 4C  8E 54 14 DC   [msg]
After CAM: 7D 6C 8A BF  AD 68 48 D8  C5 FB CD 1E  AF F2 44 99
After xor: DC 49 77 2A  0D BA 75 D8  C5 FB CD 1E  AF F2 44 99   [msg]
After CAM: 19 99 AB 92  5E 30 46 96  3D EF FB 1B  4C 87 F7 76
MIC tag  : 19 99 AB 92  5E 30 46 96  3D EF
CTR Start: 01 00 C2 9B  2C AA C4 CD  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: 02 B9 D4 1F  87 E0 60 E7  EF DE 6B 7E  D3 DE 5E D2
CTR[0002]: 61 49 31 C5  2F 34 AA 47  A3 E4 D3 2C  0B 36 41 C6
CTR[MIC ]: B9 9F C6 C5  96 7B AA 8E  1A 87
Total packet length =   41. [Encrypted]
           69 19 46 B9  CA 07 BE 87  05 B8 E1 B9  C4 9C FD 56
           CF 13 0A A6  25 1D C2 EC  C0 6C CC 50  8F E6 97 A0
           06 6D 57 C8  4B EC 18 27  68

=============== Packet Vector #20 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 2C 6B 75  95 EE 62 B1  60 B6 A3 1C  1C
Total packet length =   32. [Input (8 cleartext header octets)]
           D0 C5 4E CB  84 62 7D C4  C8 C0 88 0E  6C 63 6E 20
           09 3D D6 59  42 17 D2 E1  88 77 DB 26  4E 71 A5 CC
CBC IV in: 61 00 2C 6B  75 95 EE 62  B1 60 B6 A3  1C 1C 00 18
CBC IV out:35 A9 48 70  F9 B0 C7 85  FB 32 1A D1  3C 8C A4 9A
After xor: 35 A1 98 B5  B7 7B 43 E7  86 F6 1A D1  3C 8C A4 9A   [hdr]
After CAM: 0A 3C E3 0F  AC 09 DC 5C  00 10 5C 69  AC 19 F7 19
After xor: C2 FC 6B 01  C0 6A B2 7C  09 2D 8A 30  EE 0E 25 F8   [msg]
After CAM: 61 CD 80 D0  72 E6 84 E1  BF E1 4A 00  27 2A 4D 96
After xor: E9 BA 5B F6  3C 97 21 2D  BF E1 4A 00  27 2A 4D 96   [msg]
After CAM: E5 F9 F2 AB  47 FD 7B 8D  6F 72 F4 72  74 D7 69 BB
MIC tag  : E5 F9 F2 AB  47 FD 7B 8D  6F 72
CTR Start: 01 00 2C 6B  75 95 EE 62  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: 9C 0E 31 66  B2 81 58 31  5E 63 16 5A  9D BD CE 35
CTR[0002]: 00 3E 66 D3  E0 5F 7E A7  EF C8 9A 5F  DD 39 E3 54
CTR[MIC ]: 9A 5E 87 1A  17 10 38 0E  AA DB
Total packet length =   42. [Encrypted]
           D0 C5 4E CB  84 62 7D C4  54 CE B9 68  DE E2 36 11
           57 5E C0 03  DF AA 1C D4  88 49 BD F5  AE 2E DB 6B
           7F A7 75 B1  50 ED 43 83  C5 A9

=============== Packet Vector #21 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 C5 3C D4  C2 AA 24 B1  60 B6 A3 1C  1C
Total packet length =   33. [Input (8 cleartext header octets)]
           E2 85 E0 E4  80 8C DA 3D  F7 5D AA 07  10 C4 E6 42
           97 79 4D C2  B7 D2 A2 07  57 B1 AA 4E  44 80 02 FF
           AB
CBC IV in: 61 00 C5 3C  D4 C2 AA 24  B1 60 B6 A3  1C 1C 00 19
CBC IV out:2A 3C 23 B2  43 F5 1C 35  F7 79 5A CB  3B 20 21 2F
After xor: 2A 34 C1 37  A3 11 9C B9  2D 44 5A CB  3B 20 21 2F   [hdr]
After CAM: A1 7E AD 4C  EE AB 51 21  1D 2A 32 F2  D4 45 A6 D6
After xor: 56 23 07 4B  FE 6F B7 63  8A 53 7F 30  63 97 04 D1   [msg]
After CAM: A9 A1 32 55  8F C6 9B 98  A9 CC 23 96  FE CA 84 EB
After xor: FE 10 98 1B  CB 46 99 67  02 CC 23 96  FE CA 84 EB   [msg]
After CAM: 6A 5E 04 42  D1 A5 7E 17  9A 6C 8B 56  F7 19 80 C5
MIC tag  : 6A 5E 04 42  D1 A5 7E 17  9A 6C
CTR Start: 01 00 C5 3C  D4 C2 AA 24  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: 46 1D EF 41  AF A2 94 52  5D 51 AE CB  04 49 74 CD
CTR[0002]: 29 2E 62 66  1B 66 9A 2B  97 72 6B 77  32 A8 DC 35
CTR[MIC ]: B8 54 06 A2  6C 6F 93 37  8A BF
Total packet length =   43. [Encrypted]
           E2 85 E0 E4  80 8C DA 3D  B1 40 45 46  BF 66 72 10
           CA 28 E3 09  B3 9B D6 CA  7E 9F C8 28  5F E6 98 D4
           3C D2 0A 02  E0 BD CA ED  20 10 D3

=============== Packet Vector #22 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 BE E9 26  7F BA DC B1  60 B6 A3 1C  1C
Total packet length =   31. [Input (12 cleartext header octets)]
           6C AE F9 94  11 41 57 0D  7C 81 34 05  C2 38 82 2F
           AC 5F 98 FF  92 94 05 B0  AD 12 7A 4E  41 85 4E
CBC IV in: 61 00 BE E9  26 7F BA DC  B1 60 B6 A3  1C 1C 00 13
CBC IV out:20 60 6A D1  E1 A0 84 52  2F A3 8B F4  88 1D D6 8B
After xor: 20 6C 06 7F  18 34 95 13  78 AE F7 75  BC 18 D6 8B   [hdr]
After CAM: 71 FD FF E7  D9 C8 95 75  D3 EC 0B 7E  7B 8B BE E7
After xor: B3 C5 7D C8  75 97 0D 8A  41 78 0E CE  D6 99 C4 A9   [msg]
After CAM: CA AD 93 9C  59 BA 40 AA  1A 0B 88 1B  EE 3D 3C 65
After xor: 8B 28 DD 9C  59 BA 40 AA  1A 0B 88 1B  EE 3D 3C 65   [msg]
After CAM: DC 48 8F AA  9C 75 E7 03  17 56 C2 C7  48 48 8D 1B
MIC tag  : DC 48 8F AA  9C 75 E7 03  17 56
CTR Start: 01 00 BE E9  26 7F BA DC  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: 56 F0 17 B3  BD 09 02 D6  EA A5 A2 91  AD 4A 2D E5
CTR[0002]: 20 3D 34 21  EF 5B F8 FC  7B 21 5C 76  7B A5 21 A6
CTR[MIC ]: F1 A2 86 9C  2A 9E B8 61  48 0B
Total packet length =   41. [Encrypted]
           6C AE F9 94  11 41 57 0D  7C 81 34 05  94 C8 95 9C
           11 56 9A 29  78 31 A7 21  00 58 57 AB  61 B8 7A 2D
           EA 09 36 B6  EB 5F 62 5F  5D

=============== Packet Vector #23 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 DF A8 B1  24 50 07 B1  60 B6 A3 1C  1C
Total packet length =   32. [Input (12 cleartext header octets)]
           36 A5 2C F1  6B 19 A2 03  7A B7 01 1E  4D BF 3E 77
           4A D2 45 E5  D5 89 1F 9D  1C 32 A0 AE  02 2C 85 D7
CBC IV in: 61 00 DF A8  B1 24 50 07  B1 60 B6 A3  1C 1C 00 14
CBC IV out:78 FD B6 AF  61 9E 1C 8D  82 41 17 A8  73 60 1B 70
After xor: 78 F1 80 0A  4D 6F 77 94  20 42 6D 1F  72 7E 1B 70   [hdr]
After CAM: 62 2E 28 65  92 43 DB 82  88 79 09 1E  A7 24 54 67
After xor: 2F 91 16 12  D8 91 9E 67  5D F0 16 83  BB 16 F4 C9   [msg]
After CAM: 95 0E 52 08  FF 16 70 8C  1E D9 BB 06  3E 1E 41 CF
After xor: 97 22 D7 DF  FF 16 70 8C  1E D9 BB 06  3E 1E 41 CF   [msg]
After CAM: BA CD 51 FC  77 F4 02 8D  47 D5 7D 54  7D 46 33 4B
MIC tag  : BA CD 51 FC  77 F4 02 8D  47 D5
CTR Start: 01 00 DF A8  B1 24 50 07  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: 15 D6 DD DD  98 96 39 91  35 75 1A 64  B8 D8 D4 F9
CTR[0002]: 7D 61 6D 1D  EB 92 00 2B  6F FA AB 53  BC AF 69 89
CTR[MIC ]: 33 E9 27 BE  E1 59 06 9C  DB 32
Total packet length =   42. [Encrypted]
           36 A5 2C F1  6B 19 A2 03  7A B7 01 1E  58 69 E3 AA
           D2 44 7C 74  E0 FC 05 F9  A4 EA 74 57  7F 4D E8 CA
           89 24 76 42  96 AD 04 11  9C E7

=============== Packet Vector #24 ==================
CAM Key:   D7 5C 27 78  07 8C A9 3D  97 1F 96 FD  E7 20 F4 CD
Nonce =    00 3B 8F D8  D3 A9 37 B1  60 B6 A3 1C  1C
Total packet length =   33. [Input (12 cleartext header octets)]
           A4 D4 99 F7  84 19 72 8C  19 17 8B 0C  9D C9 ED AE
           2F F5 DF 86  36 E8 C6 DE  0E ED 55 F7  86 7E 33 33
           7D
CBC IV in: 61 00 3B 8F  D8 D3 A9 37  B1 60 B6 A3  1C 1C 00 15
CBC IV out:84 E6 CF DD  6A 37 68 5D  E6 71 AD 54  B3 BE FE B9
After xor: 84 EA 6B 09  F3 C0 EC 44  94 FD B4 43  38 B2 FE B9   [hdr]
After CAM: C5 0F A0 62  20 18 F1 21  0E BC 3D 2E  47 B7 B8 C3
After xor: 58 C6 4D CC  0F ED 2E A7  38 54 FB F0  49 5A ED 34   [msg]
After CAM: C4 6F 6D C3  17 3C 2A 7A  81 FC 2D DA  7F B7 C6 60
After xor: 42 11 5E F0  6A 3C 2A 7A  81 FC 2D DA  7F B7 C6 60   [msg]
After CAM: DF AB 2E 76  B0 67 50 B3  7C DD 9A AC  F3 79 17 71
MIC tag  : DF AB 2E 76  B0 67 50 B3  7C DD
CTR Start: 01 00 3B 8F  D8 D3 A9 37  B1 60 B6 A3  1C 1C 00 01
CTR[0001]: D6 D0 6C F8  16 CE D0 F1  A0 E0 AC 71  BA B9 AD 34
CTR[0002]: 76 4A FF 9A  1B F8 55 1F  68 54 39 0A  EE 37 24 28
CTR[MIC ]: 4B F4 31 B8  17 86 4B 5D  16 F2
Total packet length =   43. [Encrypted]
           A4 D4 99 F7  84 19 72 8C  19 17 8B 0C  4B 19 81 56
           39 3B 0F 77  96 08 6A AF  B4 54 F8 C3  F0 34 CC A9
           66 94 5F 1F  CE A7 E1 1B  EE 6A 2F

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5.  Security Considerations

Camellia-CTR and Camellia-CCM employs counter (CTR) mode for confidentiality. If a counter value is ever used for more that one packet with the same key, then the same key stream will be used to encrypt both packets, and the confidentiality guarantees are voided.

What happens if the encryptor XORs the same key stream with two different packet plaintexts? Suppose two packets are defined by two plaintext byte sequences P1, P2, P3 and Q1, Q2, Q3, then both are encrypted with key stream K1, K2, K3. The two corresponding ciphertexts are: 

      (P1 XOR K1), (P2 XOR K2), (P3 XOR K3)

      (Q1 XOR K1), (Q2 XOR K2), (Q3 XOR K3)

If both of these two ciphertext streams are exposed to an attacker, then a catastrophic failure of confidentiality results, because: 

      (P1 XOR K1) XOR (Q1 XOR K1) = P1 XOR Q1
      (P2 XOR K2) XOR (Q2 XOR K2) = P2 XOR Q2
      (P3 XOR K3) XOR (Q3 XOR K3) = P3 XOR Q3

Once the attacker obtains the two plaintexts XORed together, it is relatively straightforward to separate them. Thus, using any stream cipher, including Camellia-CTR, to encrypt two plaintexts under the same key stream leaks the plaintext.


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6.  IANA Considerations

There are no IANA assignments to be performed.


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7.  Acknowledgments

This document includes text borrowed from RFC 3610.


 TOC 

8.  References


 TOC 

8.1. Normative

[1] Matsui, M., Nakajima, J., and S. Moriai, “A Description of the Camellia Encryption Algorithm,” RFC 3713, April 2004 (TXT).
[2] Bradner, S., “Key words for use in RFCs to Indicate Requirement Levels,” BCP 14, RFC 2119, March 1997 (TXT, HTML, XML).

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8.2. Informative

[3] National Institute of Standards and Technology, “Advanced Encryption Standard (AES),” FIPS PUB 197, November 2001.
[4] Kato, A., Moriai, S., and M. Kanda, “The Camellia Cipher Algorithm and Its Use With IPsec,” RFC 4312, December 2005 (TXT).
[5] Moriai, S., Kato, A., and M. Kanda, “Addition of Camellia Cipher Suites to Transport Layer Security (TLS),” RFC 4132, July 2005 (TXT).
[6] Moriai, S. and A. Kato, “Use of the Camellia Encryption Algorithm in Cryptographic Message Syntax (CMS),” RFC 3657, January 2004 (TXT).
[7] Eastlake, D., “Additional XML Security Uniform Resource Identifiers (URIs),” RFC 4051, April 2005 (TXT).
[8] International Organization for Standardization, “Information technology - Security techniques - Encryption algorithms - Part 3: Block ciphers,” ISO/IEC 18033-3, July 2005.
[9] “The NESSIE project (New European Schemes for Signatures, Integrity and Encryption).”
[10] Information-technology Promotion Agency (IPA), “Cryptography Research and Evaluation Committees” (HTML).
[11] Dworkin, M., “Recommendation for Block Cipher Modes of Operation - Methods and Techniques,” NIST Special Publication 800-38A, November 2001.
[12] Whiting, D., Housley, R., and N. Ferguson, “Counter with CBC-MAC (CCM),” RFC 3610, September 2003 (TXT).
[13] National Institute of Standards and Technology, “Computer Data Authentication,” FIPS PUB 113, May 1985.

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Authors' Addresses

  Akihiro Kato
  NTT Software Corporation
Phone:  +81-45-212-7577
Fax:  +81-45-212-7800
Email:  akato@po.ntts.co.jp
  
  Masayuki Kanda
  Nippon Telegraph and Telephone Corporation
Phone:  +81-422-59-3456
Fax:  +81-422-59-4015
Email:  kanda.masayuki@lab.ntt.co.jp

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