# Useless admin
__PROBLEM__
One member of our team, piece of sh#t an#s_boss, made a huge mistake using multi time pad.
He knew that the OTP, if it is well applied, it is unbreakable. But this usesless brainless retard, went yoloooo.
Not only that, but he deleted the original plain text, so we are screwed.
The only thing left is the following file : cipher.json
It contains multiple cipher text, created using OTP and the same key.
So can you figure out the plain text of the cipher_flag ?
Flag format : Securinets{plain text here}
Author:BlueWhale
[cipher.json](https://www.ctfsecurinets.com/files/bbfe31808617c8c67fd9da96f1daa8bd/cipher.json)
__SOLUTION__
Okay so this is like super xor attack or something :smile:
The best part about this is that I didn't had to do anything because this is common know problem and someone already have created a great [script](https://gist.github.com/intrd/a12e1bb564b1825e864efa2ebeb37874)
I just had to make few changes to it.
First this is what the json data looked.
```json➜ http https://www.ctfsecurinets.com/files/bbfe31808617c8c67fd9da96f1daa8bd/cipher.json | jq .{  "cipher_list": [    "1b0605000e14000d1b524802190b410700170e10054c11480807001806004e4f1f4f01480d411400531158141e1c100016535a480c000c031a000a160d421e004113010f13451e0c0100100a020a1a4e165f500d0c1e041a090b001d0515521c0a0410000a4f4b4d1d1c184d071600071c0a521d1706540940",    "1e10524e001f11481c010010070b13024f0704590903094d0c000e4f0711000615001911454217161a1a45040149000a5218404f1e0012060b1b590a1048171741140c01174c0d49174f0c8d4fc7520211531b0b0c1e4f",    "1d0c04451352001a000154431b014109450a0a0b000045490403520a1d16490008535848085942071c0d0c57101c0045111c40430c4e111c0b1b1c451d4f071712010508475518061d00060a1b0a1a4c165d",    "160d074300061d071b524e06190b134e450a0b0a4d4c12411d004f014045491b4649074804001100011d4504520612451e165d53064e164e1d060d0d44541a0041031b0b06540d1a070004001d4b074800531c04101d4f",    "1a1d524912521548120045021b4e1506490a0859150345531d12521b4e094909030003011148420453074d161e05540b071e4c451b000a084a1d1c04084c0b45060b060a4742070618534218070210484512020043100e191e5956111a1c001c1f0b5c",    "1a1d5248000154041a1c47430d0b04000005015900140c4f04534f094e08490103000000045442111b11001b1b1d000917535a48004e021d4a0e0b0044491c03080a001a024c11490748074f02040054451a1d150c1b150d020d0e",    "1a1d5249125215481613500a1b0f0d4e4d0d1c0d000700001d1c001b06004f1d0f5a11480745040a011100181c0c540d13000e44085404404a061716014e010c0308104e084e0d4911450506011853540a5304120a1a154c0a1843001b45541c481607051b431f480d001e0400000c531d01011d00124441010200190d0800000000000e54060001100a1b4d0b040d105347",    "0a0607000913020d551300041d0f0f0a0003061f154c034f1b53530602004e0c030c541f0454110a1d5a001e0649190419165d00104f104e1b1a101101001b0b1705051b0642040c5341114f0e4b104f0803110b0a060f42",    "160d074300061d071b524e06190b134e450a0b0a4d4c12411d004f014045491b4649074804001100011d4504520612451e165d53064e16424a1810110c00060d04440e1c02411c0c00544209001953540d165009021a1542",    "1e10524e001f11481c010010070b13024f0704590903094d0c000e4f0711000615001911454217161a1a45040149000a5218404f1e0012060b1b590a1048171741140c01174c0d49174f4201001f534b0b1c074b",    "1a49134d4113540a0713490d434e160f541700174f4c11480c53520a1d1100000000190d4549114512544d12000c540402034b4e0d491d40"  ],  "cipher_flag": "1a4905410f06110c55064f430a00054e540c0a591603174c0d5f000d1b110006414c1848164516111f1100111d1b54001c17474e0e001c011f1d0a4b"}```
After using the [cipher.py](cipher.py) we get the output:
```bash➜ python2 cipher.pyFix this sentence:* *anted to ind t** y*rl*, but i'*l*settle for endin* ****s.
Fixed:cure let me be know a
('key is', 'y