Two players, S and T, are playing a game where they make alternate moves. S plays first.
In this game, they start with an integer N. In each move, a player removes one digit from the integer and passes the resulting number to the other player. The game continues in this fashion until a player finds he/she has no digit to remove when that player is declared as the loser.

With this restriction, it’s obvious that if the number of digits in N is odd then S wins otherwise T wins. To make the game more interesting, we apply one additional constraint. A player can remove a particular digit if the sum of digits of the resulting number is a multiple of 3 or there are no digits left.

Suppose N = 1234. S has 4 possible moves. That is, he can remove 1, 2, 3, or 4. Of these, two of them are valid moves.

- Removal of 4 results in 123 and the sum of digits = 1 + 2 + 3 = 6; 6 is a multiple of 3.
- Removal of 1 results in 234 and the sum of digits = 2 + 3 + 4 = 9; 9 is a multiple of 3.
The other two moves are invalid.

If both players play perfectly, who wins?

Input
The first line of input is an integer T(T<60) that determines the number of test cases. Each case is a line that contains a positive integer N. N has at most 1000 digits and does not contain any zeros.

Output
For each case, output the case number starting from 1. If S wins then output ‘S’ otherwise output ‘T’.

Sample Input Output for Sample Input

题目大意：两个人博弈，轮流从一列数字里取数，要求剩下的数字之和必须是3的倍数，或者为空。不满足条件者输。

比较简单的博弈，两种情况，一种情况只有一个数字，后手输。第二种情况又分两种情况，第一种是先手可以取数，那么取完之后，接下来的人必取3的倍数，只要统计出三的倍数有多少个就ok了。

#include
  
   
#include
   
     char a[1005]; int b[1005]; int main() { int t; scanf("%d%*c",&t); for(int ca=1;ca<=t;ca++) { scanf("%s",a); printf("Case %d: ",ca); int n=strlen(a); if(n == 1) { printf("S\n"); continue; } int cnt=0 , sum=0 ,cnt2=0; for(int i=0;i
    
     
 
     
 
     

 
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