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Does adding complexity mean a more secure cipher?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Easy explanation of “IND-” security notions?Is this hand cipher any more secure than the Vigenère cipher?What does ''latency'' really mean when a block cipher is partially unrolled?RSA: how does it work and how is it more secure than symmetric systemsComplexity of attacks on affine cipherIs my protocol that uses hybrid cryptography and AES-GCM secure?What does 0…0 and 1…1 meanDoes AAD make GCM encryption more secure?What does (block cipher) decryption parallelizable mean?What does “$<!!<!!<$” mean?What does fullstop mean in this context?
$begingroup$
I have a cryptography workshop question I'm having trouble with as follows;
Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.
Person B recommends "increasing" security of the cipher by instead doing :
$(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$
Does this in fact increase security of the cipher or increase new problems.
My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.
Unfortunately the above context is all I have been provided for this question.
encryption
$endgroup$
migrated from stackoverflow.com Apr 11 at 14:18
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
I have a cryptography workshop question I'm having trouble with as follows;
Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.
Person B recommends "increasing" security of the cipher by instead doing :
$(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$
Does this in fact increase security of the cipher or increase new problems.
My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.
Unfortunately the above context is all I have been provided for this question.
encryption
$endgroup$
migrated from stackoverflow.com Apr 11 at 14:18
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
I have a cryptography workshop question I'm having trouble with as follows;
Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.
Person B recommends "increasing" security of the cipher by instead doing :
$(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$
Does this in fact increase security of the cipher or increase new problems.
My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.
Unfortunately the above context is all I have been provided for this question.
encryption
$endgroup$
I have a cryptography workshop question I'm having trouble with as follows;
Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.
Person B recommends "increasing" security of the cipher by instead doing :
$(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$
Does this in fact increase security of the cipher or increase new problems.
My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.
Unfortunately the above context is all I have been provided for this question.
encryption
encryption
edited Apr 11 at 14:59
Ella Rose♦
17.1k44585
17.1k44585
asked Apr 11 at 14:08
melloncolliemelloncollie
417
417
migrated from stackoverflow.com Apr 11 at 14:18
This question came from our site for professional and enthusiast programmers.
migrated from stackoverflow.com Apr 11 at 14:18
This question came from our site for professional and enthusiast programmers.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
$endgroup$
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
Apr 11 at 14:59
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
Apr 11 at 15:01
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
Apr 11 at 15:14
add a comment |
$begingroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
$endgroup$
add a comment |
$begingroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
$endgroup$
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
Apr 11 at 14:59
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
Apr 11 at 15:01
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
Apr 11 at 15:14
add a comment |
$begingroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
$endgroup$
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
Apr 11 at 14:59
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
Apr 11 at 15:01
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
Apr 11 at 15:14
add a comment |
$begingroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
$endgroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_texta = E_k(m) oplus m\c_textb = E_k(m) oplus 1111 dots 11\c' = c_textb oplus 1111dots11\m = c_texta oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
edited Apr 11 at 15:12
answered Apr 11 at 14:47
Ella Rose♦Ella Rose
17.1k44585
17.1k44585
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
Apr 11 at 14:59
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
Apr 11 at 15:01
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
Apr 11 at 15:14
add a comment |
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
Apr 11 at 14:59
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
Apr 11 at 15:01
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
Apr 11 at 15:14
2
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
Apr 11 at 14:59
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
Apr 11 at 14:59
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
Apr 11 at 15:01
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
Apr 11 at 15:01
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
Apr 11 at 15:14
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
Apr 11 at 15:14
add a comment |
$begingroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
$endgroup$
add a comment |
$begingroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
$endgroup$
add a comment |
$begingroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
$endgroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
answered Apr 11 at 15:06
Marc IlungaMarc Ilunga
42718
42718
add a comment |
add a comment |
$begingroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
$endgroup$
add a comment |
$begingroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
$endgroup$
add a comment |
$begingroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
$endgroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
answered Apr 11 at 23:32
MarkMark
1885
1885
add a comment |
add a comment |
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