#include<stdio.h>
#include<stdlib.h>
int gys(int,int);

int main()
{
    int c,m,n,i,j,a,length;
    for(scanf("%d",&c);c;c--)
    {
         scanf("%d%d",&m,&n);
         int num1a[m][n],num1b[m][n],num2a[m*n],num2b[m*n];
         for(i=1;i<=m;i++)
         {
            for(j=1;j<=n;j++)
            {
                  a=gys(i,j);
                  num1a[i-1][j-1]=i/a;
                  num1b[i-1][j-1]=j/a;         
                       }               
                        }
         num2a[0]=num1a[0][0];
         num2b[0]=num1b[0][0];
         length=1;   
         for(i=0;i<m;i++)
         {
                for(j=0;j<n;j++)
                {
                                for(a=0;a<length;a++)
                                   {
                                   if((num1a[i][j]==num2a[a])&&(num1b[i][j]==num2b[a]))
                                   break;
                                   }
                                if(a==length)
                                {
                                   num2a[length]=num1a[i][j];
                                   num2b[length]=num1b[i][j];
                                   length++;
                                             }
                                }
                                
                         }  
          printf("%d\n",length);            
          }
}
int gys(int m,int n){
    int r=m%n;
    while(r!=0)  
    {  
     m=n;  
     n=r;  
     r=m%n;  
           }  
     return n;
}

平面上有m*n个整点，他们的坐标(x, y)满足1<=x<=m, 1<=y<=n, x,y都是整数。求从原点能看到的点的数量（如果某点与原点的连线上没有其他点，则该点能被原点看到）。
Input
第一行一个数t（1<=t<=15），表示数据的组数
以下每组数据一行，每行两个数m,n（0<=m, n<=50000）
Output
对于每组数据，输出能被原点看到的点的总数