A frog has just learned some number theory, and can't wait to show his ability to his girlfriend.

Now the frog is sitting on a grid map of infinite rows and columns. Rows are numbered 1,2,⋯from the bottom, so are the columns. At first the frog is sitting at grid (sx,sy), and begins his journey.

To show his girlfriend his talents in math, he uses a special way of jump. If currently the frog is at the grid (x,y), first of all, he will find the minimum z that can be divided by both x and y, and jump exactly z steps to the up, or to the right. So the next possible grid will be (x+z,y), or (x,y+z)

After a finite number of steps (perhaps zero), he finally finishes at grid (ex,ey). However, he is too tired and he forgets the position of his starting grid!

It will be too stupid to check each grid one by one, so please tell the frog the number of possible starting grids that can reach (ex,ey)!

3

6 10

6 8

2 8

Case #1: 1

Case #2: 2

Case #3: 3

【题目大意】：一只青蛙在坐标系中跳跃，假设它某一时刻位于（x,y）,那么它下一次可以跳到（x+z，y） 或者 （x, y+z），其中Z = LCM（x , y），就是x,y的最小公倍数。现在已知青蛙跳到了（ex,ey）
问青蛙可能的起点有多少个？从这样的起点跳若干次可以到达(ex , ey),可以一次不跳，要跳必须向右跳或者向上跳，步长为z。

【题解】：首先明白这样一个事实，假设能到达终点（ex,ey）的最左下角的起点是（x,y）， 那么沿途的所有点（x',y'）都是可以到达终点的点。 设GCD（ex , ey） = k, 由于每次跳跃的步长是原坐标中横坐标x和纵坐标y的LCM，即z，从（x,y）出发，不论新的坐标（ x' , y'）是（x+z，y） 还是 （x, y+z），GCD（x' , y'）始终等于k
证明如下：不妨下一步跳到（x+z,y） 设gcd(x,y) = s ; x = ms, y = ns; 显然mn互质
下一步坐标（x',y;） gcd(x',y') = gcd(ms + (ms * ns) / s , ns) = gcd( m*(1+n) , n)
由于m和n互质， n+1 和 n也互质，所以m*(1+n) 和 n必然互质，所以gcd(x',y') = s所以沿途每一个点的坐标x,y的gcd都相等，等于什么呢，等于最后终点的gcd(ex,ey) = k，这是给定的

【递推】
正推不好推，可以从终点反推，递推公式 f(x , y) = f(x-z1, y) + f(x, y-z2)，其中z1 = (x-z1) * y / k  即z1 = x*y / (k+y)
同理 z2 = x*y / (k+x)

【代码】f函数可以写成循环的，若写成递归的，更新x，y坐标的操作一定要写在递归函数的参数里，不能在此之前做，否则会WA，原因是回溯问题

#include<iostream>

using namespace std;

int gcd(int a, int b){

    if(b == )    return a;

    else return gcd(b,a%b);

}

int k;

long long ans = ;

void  f(long long x, long long y){

    if(x == y){

        return ;

    }

    //其实要往右走x > y是必须的，自己可以稍微证明  ，加一个这个条件可以快一点不加也行

    long long  z1 = (x*y) / (k+y);

    //注意保持每一步GCD都是K

    if( x>y && (x*y) % (k+y) ==  && gcd(x-z1 , y) == k){

        ans++;

        //cout<<"往右边走 x =  "<<x<<" y = "<<y<<endl;

        f(x - z1,y);
　　　　　　//写成x = x - z1 然后 f(x,y)会WA

    }

    long long z2 = (x*y) / (k+x);

    if(y > x && (x*y) % (k+x) ==  && gcd(x , y-z2) == k){

        ans++;

        //cout<<"往上边走 x =  "<<x<<" y = "<<y<<endl;

        f(x,y - z2);

    }

} 

int main(){

    int ex,ey;

    int t;

    cin>>t;

    int cas=;

    while(t--){

        ans = ;

        cin>>ex>>ey;

        k = gcd(ex,ey);

        f(ex,ey);

        cout<<"Case #"<<cas++<<": ";

        cout<<ans+<<endl;

    }

    return ;

}