Fermat goes hacking RSA

Problem #153

Tags: popular-algorithm c-1 c-0

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This problem is a modification of the similar exercise from Stanford-online course on Cryptography.

If after solving RSA cryptography exercise you still have vague idea of security of the RSA algorithm here is a variation of the same problem.

You are again to decrypt the encoded message but now instead of p and q you are given only their product n, just as a real attacker. Encryption was still performed with e=65537, but unfortunately you have no idea of the decryption exponent d!

However by chance you know that the person who encrypted the message was a perfect noob and used close values of p and q, so you may find a way to factorize n easily enough.

Conversion between string and number is performed in the same way as in the RSA-related exercise mentioned above.

Input data will contain n in the first line.
The second line contains cipher which was generated as a^65537 mod n where a is the original text converted to long integer.
Answer should contain the deciphered text.

Example:

input data:
2005386240811006492510206908835874977464399827995998174235015291258133373258958037573585627
258926557618335589879504876460462075566410747651590614428022205934562315249635550863811428

answer:
EGG EAT SKI SHY ARM EON HIP FUN LOW

Hint: The name of Pierre de Fermat is presented in the title intentionally, so some googling may help...

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