GCPC11F  Diary
Nowadays, people who want to communicate in a secure way use asymmetric encryption algorithms such as RSA. However, my older brother uses another, simpler encryption method for his diary entries. He uses a substitution cipher where each letter in the plaintext is substituted by another letter from the alphabet. The distance between the plaintext letter and the encrypted letter is fixed. If we would define this fixed distance d to 5, A would be replaced by F, B by G, C by H, ..., Y by D, Z by E.
With a fixed and known distance d the decryption would be somewhat simple. But my brother uses random distances for each of his diary entries. To decrypt his diary I have to guess the distance d for each entry. Thus, I use the well known phenomenon that the letter E is used more often in English words than other letters. Can you write a program for me that calculates the distance d based on the fact that the most used letter in the encrypted text corresponds to the letter E in plaintext? Of course, I am interested in the decrypted text, too.
Input
The input consists of several test cases c that follow (1 ≤ c ≤ 100). Each test case is given in exactly one line containing one diary entry. Diary entries only use upper case letters (AZ) and spaces. Each diary entry consists of at most 1000 encrypted letters (including spaces).
Output
For each test case, print one line containing the smallest possible distance d (0 ≤ d ≤ 25) and the decrypted text. If the decryption is not possible because there are multiple distances conforming to the rules above, print NOT POSSIBLE instead. Spaces are not encrypted.
Example
Input: 4 RD TQIJW GWTYMJWX INFWD JSYWNJX ZXJ F XNRUQJ JSHWDUYNTS YJHMSNVZJ THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG XVIDRE TFCCVXZRKV GIFXIRDDZEX TFEKVJK UVTIPGKZFE XVIDRE TFCCVXZRKV GIFXIRDDZEX TFEKVJK Output: 5 MY OLDER BROTHERS DIARY ENTRIES USE A SIMPLE ENCRYPTION TECHNIQUE 10 JXU GKYSA RHEMD VEN ZKCFI ELUH JXU BQPO TEW 17 GERMAN COLLEGIATE PROGRAMMING CONTEST DECRYPTION NOT POSSIBLE
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atulav:
20171201 06:51:05
cake walk !!!


nadstratosfer:
20170902 19:17:02
Finally AC after ifelsing my way out of 12 char cases. 

akshayvenkat:
20160620 12:51:04
one of the most irritating problems i have ever come across :/ never knew one can brainf**k us even while using c++ ;) 

minhthai:
20160209 07:29:21
wow, simple but very easy to make mistake 

Dushyant Singh:
20150430 17:33:39
Distance cannot be negative and it is also not always less than 13. It is 0 ≤ d ≤ 25. Comments are misleading. 

Angel Gonzalez:
20141101 21:47:47
You should've specified the distance is meant to be calculated in only one direction (From right to left). Therefore distance from A to E is 22 not 4. Many silly WA's because of that. Not to mention the fact that the problem states : "calculate de min distance" leads to many misinterpretations. Last edit: 20141101 21:51:02 

Sahil Dua:
20140812 17:29:37
Easy job in Python. Time limit isn't a problem at all! Just do what strikes your brain just after reading the statement! 

jiglipufff:
20131228 07:14:33
10 attempts n finally the GREEN signal :) 

Arianto Wibowo:
20130704 03:18:25
@nikoo28


Vishal Sharma:
20130701 20:05:46
dont know why but getting wa again and again what is the output of single space or newline as input NOT POSSIBLE OR 0 
Added by:  Adrian Kuegel 
Date:  20110705 
Time limit:  0.407s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 
Resource:  German Collegiate Programming Contest 2011 (Author: Tobias Werth) 