C. Vanya and Scales

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

Copy

3 7

output

Copy

YES

input

Copy

100 99

output

Copy

YES

input

Copy

100 50

output

Copy

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.

题意:

有砝码重为w1次方,w2次方,w3次方....,问能不能称出质量为m的物品。

思路:

可以列出一个式子a1*w1+a2*w2+a3*w3+..=m;

根据题意 ai的值只能为0，1，-1.

这个式子有点类似进制转换,但关键是-1的处理其实-1可以理解为m分解成当前位有w-1个w需要加一个w。处理好这个就好写代码

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define LL long long

using namespace std;

int main()
{
    int w,m;
    cin>>w>>m;
    if(w<=2)
    {
        cout<<"YES"<<endl;
        return 0;
    }
    while(m)
    {
       if(!((m-1)%w)) m--;
       else if(!((m+1)%w)) m++;
       else if(m%w)
       {
           cout<<"NO"<<endl;
           return 0;
       }
       m/=w;
    }
    cout<<"YES"<<endl;
    return 0;
}