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

每次加MmodN，如果每次加N能遍歷整個(gè)集合的話，那他一定是mod N+剩余類的生成元。生成元的條件是M與N互質(zhì)。

import java.util.Scanner;public class Main {static int gcd(int a, int b) {if (b == 0)return a;elsereturn gcd(b, a % b);}public static void main(String[] args) {Scanner in = new Scanner(System.in);while (in.hasNext()) {int a = in.nextInt();int b = in.nextInt();if (a == -1 && b == -1) {break;}if (gcd(a, b) == 1)System.out.println("YES");else System.out.println("POOR Haha");}in.close();} }

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