FAKEHASH - Faketorial Hashing


Are you familiar with polynomial hashing? If you are not, all the better! You don’t need to know what polynomial hashing is, the world is better off without it. I hate polynomial hashing so much that I found a new way to hash strings. It is called the Faketorial Hashing.

First, let’s define a function, ord(ch) = the position of ch in the alphabet + 1, where ch can be any lowercase letter. So, ord(a) = 2, ord(b) = 3, ord(c) = 4, … ord(z) = 27.

Let fact(x) be x! or the factorial of x. A few examples, fact(1) = 1, fact(2) = 2, fact(3) = 6, fact(4) = 24, fact(5) = 120, etc. Given a string S of length N, consisting of lowercase letters only, the Faketorial Hashing of S, is defined as below:

fake_hash(S) = fact(ord(S[0])) × fact(ord(S[1])) × fact(ord(S[2])) × …… × fact(ord(S[N - 1]))

In other words, it is the product of the factorial of the ord() value of all the characters in S (That’s right, no modulus! Unlike the lame polynomial hashing). Not only that we have a new hashing mechanism in place, but we would also like to crack this now. Given a string S1 consisting of lowercase letters only, your task is to find a different string S2 consisting of lowercase letters, such that, fake_hash(S1) = fake_hash(S2) and S1 ≠ S2.

If there are multiple possible choices for S2, you need to find the lexicographically smallest one, or output the word “Impossible” without quotes, if it is not possible to find such a string.

Input

The first line contains an integer T, denoting the number of test cases. Each test case contains the string S1 consisting of lowercase letters (a-z) only.

Constraints

  • 1 ≤ T ≤ 3000
  • 1 ≤ |S1| ≤ 30


  • Except for the sample, the following constraints will hold:
  • 1 ≤ |S1| ≤ 5, for 90% of the test cases
  • 1 ≤ |S1| ≤ 15, for 99% of the test cases
  • Output

    For each test case, output the case number followed by the required output. Please refer to the sample input/output section for the precise format.

    Example

    Input:
    10
    tourist
    petr
    mnbvmar
    bmerry
    xellos
    sevenkplus
    dragoon
    zzz
    snapdragon
    zosovoghisktwnopqrstuvwxyzoos
    
    
    Output:
    Case 1: aaaaabbdnstttu
    Case 2: aqst
    Case 3: abmmnrv
    Case 4: aaabbnrry
    Case 5: aaaaaaadddlnuz
    Case 6: aaaaaaabbddddnquuuz
    Case 7: aaaaaaaaaaaaabdnnnt
    Case 8: Impossible
    Case 9: aaaaaaaaaabdnnnpst
    Case 10: aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaffffjnnnnqqttttuuuuxxzzzzzz
    
    

    hide comments
    nogor: 2019-10-17 03:47:50

    Is it doable with Naive Backtracking? I need help.

    ANS: It is definitely doable with backtracking, but the naive approach won't work. You need some observations and some pruning.

    Last edit: 2019-11-12 20:45:35
    wisfaq: 2019-01-24 10:35:48

    For some reason the answer checker keeps running for days (or so it seems) without coming with a verdict.

    ANS: Thanks for noticing. Not sure what went wrong. My previously accepted solutions are now running indefinitely. Anyway, I re-uploaded the i/o and it seems to be working now.

    Last edit: 2019-02-02 19:36:58

    Added by:sgtlaugh
    Date:2019-01-19
    Time limit:12s
    Source limit:50000B
    Memory limit:1536MB
    Cluster: Cube (Intel G860)
    Languages:All
    Resource:Own Problem, Used in 2018-2019 ACM-ICPC Asia Dhaka Regional Contest