GG and MM like playing a game since they are children. At the beginning of game, there are two piles of stones. MM chooses a pile of stones first, which has x stones, and then she can choose a positive number k and remove k*x stones out from the other pile of stones, which has y stones (I think all of you know that y>=k*x - -!). Then it comes the turn of GG, followed the rules above-mentioned as well. When someone can't remove any stone, then he/she loses the game, and this game is finished. 
Many years later, GG and MM find this game is too simple, so they decided to play N games at one time for fun. MM plays first, as the same, and the one on his/her turn must play every unfinished game. Rules to remove are as same as above, and if someone cannot remove any stone (i.e., loses the last ending game), then he/she loses. Of course we can assume GG and MM are clever enough, and GG will not lose intentionally, O(∩_∩)O~ 
InputThe input file contains multiply test cases (no more than 100). 
The first line of each test case is an integer N, N<=1000, which represents there are N games, then N lines following, each line has two numbers: p and q, standing for the number of the two piles of stones of each game, p, q<=1000(it seems that they are so leisure = =!), which represent the numbers of two piles of stones of every game. 
The input will end with EOF. 
OutputFor each test case, output the name of the winner. 
Sample Input

3
1 1
1 1
1 1
1
3 2

Sample Output

MM
GG

題意：

每組給n個遊戲，每個遊戲有兩堆石頭，GG和MM輪流操作，操作規則：

從兩堆裡面選出一堆，假設這堆石頭有x個，然後在另一堆裡取k*x個石頭(k是正整數)

誰不能取石頭誰輸，MM先手。

思路：

這是一個every——sg遊戲。

決定總遊戲勝負的是最後一局遊戲的勝負。因為不能取石頭的情況就已經是最後一局了，所以之前的遊戲勝負情況沒有意義。

那麼為了自己能贏，對於自己會贏的遊戲，我肯定想盡可能地延長時間，對於自己會輸的遊戲，我肯定想盡可能地結束。

那麼可以找出每一局所走的時間，最後進行判斷即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <stack>
#define INF 0x3f3f3f3f
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int n;
int cnt;
int dfs(int a, int b)  //計算步數
{
    if(a<b)
    {
        a^=b;
        b^=a;
        a^=b;
    }
    if(b==0)
        return 0;
    if(!dfs(b, a%b))  //如果可以把輸的局面留給對手，那麼我就是贏
    {
        cnt++;
        return 1;
    }
    if(a/b>1)    //如果a/b>1,那麼勝負也決定在我手中
    {            //因為對於a%b這局面只可能是贏或輸,我就可以對應地不留或留b個，使局面對我有利
        cnt+=2;
        return 1;
    }
    cnt++;
    return 0;
}
int main()
{
    while(~scanf("%d", &n))
    {
        int a=0,b=0;   //a表示MM能贏的遊戲中步數最大值,b是GG能贏的最大步數
        for(int i=0; i<n; i++)
        {
            int x, y;
            scanf("%d%d", &x, &y);
            cnt=0;
            if(dfs(x, y))
                a=max(cnt, a);
            else
                b=max(b, cnt);
        }
        printf("%s\n", a>b?"MM":"GG");
    }
    return 0;
}