/*
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, its 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?
*/

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>

using namespace std;

bool prob(string &str1)
{
	string str2 = str1;
	string str3 = str2;
	if (str1 == "") return false;
	for(int i=0; i<str2.size(); i++)
	{
		int counter = 0;
		str2.erase(i, 1);
		for (int j = 0; j < str2.size(); j++)
			counter = counter + str2[j] - '0';
		if (counter % 3 == 0) {
			str1 = str2; return true;
		}
		else str2 = str1;
	}
	return false;
}

int main()
{
	int testcases; cin >> testcases;
	for (int i = 0; i < testcases; i++)
	{
		cout << "Case " << i+1 << ":";
		int arr[2] = { 0,1 };
		string str; cin >> str;
		while (prob(str))
		{
			arr[0] = 1 - arr[0]; arr[1] = 1 - arr[1];
		}
		if (arr[0] == 1)
			cout << "S\n";
		else cout << "T\n";
	}


	return 0;

}