Codeforces 552C. Vanya and Scales【巧】

C. Vanya and Scales

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 mass m 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.

Examples

input

3 7

output

YES

input

100 99

output

YES

input

100 50

output

NO

Note

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.

恩，砝码，放天平上，问是否能使天平平衡，天平两侧都可放砝码，每个砝码只能用一次。比较巧妙的想法，将m 换成w进制，因为砝码w^0 ,w^1 ,w^2 ,w^3等等即1 ，10 ，100,1000等等，所以只有换成w 进制之后某位数值为w-1才可加1 ，而为0 或1 的不用处理，为其他数值则要输出NO 

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int r[1010];int main(){    int w,m;    while(~scanf("%d%d",&w,&m))    {        int k=0;        while(m)        {            r[k++]=m%w;            m=m/w;        }        int i;        for(i=0;i<k;++i)        {            if(r[i]>=w)//开始没想到处理这种情况，看错误数据，调试一遍才找到            {                r[i+1]++;                r[i]=r[i]%w;            }            if(r[i]<=1)                continue;            else if(r[i]==w-1)                r[i]=0,r[i+1]++;            else                break;        }        if(i==k)            printf("YES\n");        else            printf("NO\n");    }    return 0;}