CodeForces 552C Vanya and Scales

Vanya and Scales

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 552C

Description

Vanya has a scales for weighing loads and weights of masses w⁰, w¹, w², ..., w¹⁰⁰ grams where w is some integer not less than 2(exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of massm and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.

Input

The first line contains two integers w, m (2 ≤ w ≤ 10⁹, 1 ≤ m ≤ 10⁹) — the number defining the masses of the weights and the mass of the item.

Output

Print word 'YES' if the item can be weighted and 'NO' if it cannot.

Sample Input

Input

3 7

Output

YES

Input

100 99

Output

YES

Input

100 50

Output

NO

Hint

Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.

Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.

Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.

 #include <stdio.h>
 #include <string.h>
 int main()
 {
     int w,m;
     int i,j,flg;
     while(scanf("%d %d",&w,&m)!=EOF)
     {
         flg=;
         while(m)
         {
             int a=m%w;
             if(a== || a==)
             {
                 m=m/w;
             }
             else if(a==w-)
             {
                 m=m/w+;
             }
             else
             {
                 flg=;
                 break;
             }
         }
         if(flg)
             printf("YES\n");
         else
             printf("NO\n");
     }
     return ;
 }