Problem Description
The Children’s Day has passed for some days .Has you remembered something happened at your childhood? I remembered I often played a game called hide handkerchief with my friends.
Now I introduce the game to you. Suppose there are N people played the game ,who sit on the ground forming a circle ,everyone owns a box behind them .Also there is a beautiful handkerchief hid in a box which is one of the boxes .
Then Haha(a friend of mine) is called to find the handkerchief. But he has a strange habit. Each time he will search the next box which is separated by M-1 boxes from the current box. For example, there are three boxes named A,B,C, and now Haha is at place of A. now he decide the M if equal to 2, so he will search A first, then he will search the C box, for C is separated by 2-1 = 1 box B from the current box A . Then he will search the box B ,then he will search the box A.
So after three times he establishes that he can find the beautiful handkerchief. Now I will give you N and M, can you tell me that Haha is able to find the handkerchief or not. If he can, you should tell me "YES", else tell me "POOR Haha".
 
Input
There will be several test cases; each case input contains two integers N and M, which satisfy the relationship: 1<=M<=100000000 and 3<=N<=100000000. When N=-1 and M=-1 means the end of input case, and you should not process the data.
 
Output
For each input case, you should only the result that Haha can find the handkerchief or not.
 
Sample Input
3 2
-1 -1
 
Sample Output
YES
 

当从m开始找时，如果m与n有公约数时，在找盒子时，就只会在这些约数的倍数之间找，而不是约数倍数的盒子就永远也不会被找到，所以当m与n的最大公约数为1，即m与n互质时才能找遍所有的盒子。

(一)

#include<stdio.h>
#include<math.h>
int Zhisu(int a,int b)
{
    if((a-b)==0)
        return b;
    else
        Zhisu(b,abs(a-b));
}
void main()
{
    int a,b;
    scanf("%d %d",&a,&b);
    while(a!=-1 && b!=-1)
    {
        if(Zhisu(a,b)==1)
            printf("YES\n");
        else
            printf("POOR Haha\n");
        scanf("%d %d",&a,&b);
    }
}

(二)

#include<stdio.h>
void main()
{
    int a,b,c;
    while(1)
    {
        scanf("%d %d",&a,&b);
        if(a==-1 && b==-1)
            return;
        while(b!=0)
        {
            c=a%b;
            a=b;
            b=c;
        }
        if(a==1)
            printf("YES\n");
        else
            printf("POOR Haha\n");
    }
}