Wolf and Rabbit

There is a hill with n holes around. The holes are signed from 0 to n-1.



A rabbit must hide in one of the holes. A wolf searches the rabbit in anticlockwise order. The first hole he get into is the one signed with 0. Then he will get into the hole every m holes. For example, m=2 and n=6, the wolf will get into the holes which are signed 0,2,4,0. If the rabbit hides in the hole which signed 1,3 or 5, she will survive. So we call these holes the safe holes.

Input

The input starts with a positive integer P which indicates the number of test cases. Then on the following P lines,each line consists 2 positive integer m and n(0<m,n<2147483648).

Output

For each input m n, if safe holes exist, you should output "YES", else output "NO" in a single line.

Sample Input

21 22 2

Sample Output

NO

YES

思路：   思考一下，每次到的位置依次是  m%n,2m%n,3m%n,4m%n...km%n  尝试一下可知 当两个数互质的时候，就可以遍历所有的洞
代码
#include<stdio.h>int gcd(int a,int b) //递归{return b==0?a:gcd(b,a%b);// }
unsigned int gcd(unsigned int a,unsigned int b) //  不用递归 即 辗转相除法
{
    int r;
    while(b>0)
    {
         r=a%b;
         a=b;
         b=r;
    }
    return a;
}
int main(){int p;scanf("%d",&p);while(p--){int n,m;scanf("%d%d",&n,&m);if(gcd(n,m)==1) printf("NO\n");else printf("YES\n");}return 0;}