Codeforces #369C Vanya and Scales

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

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.

题意：给定w和m和一个天秤。

w有w^0 w^1 w^2..w^n  看看能否从这些数中抽出一些数和m一起放，看看两边是否平衡。

做法是将m变成w进制数，如果中间出现了0 1 w-1外的数就输出NO，否则除m，m-1，m+1。

因为变成w进制0 和1可以相当于w由1和0组成所以不用考虑，而w-1可以靠w进制这位数上+1从w-1构成10所以也不予考虑。

其他的就反过来想

代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int main(){
    int m,w;
    while(cin>>w>>m){
    int a;
    int flag=1;
    while(m){
        a=m%w;
        if(a==1)
            m=(m-1)/w;
        else if(a==0)
            m=m/w;
        else if(a==w-1)
            m=(m+1)/w;
        else
           {flag=0;
            break;}
    }
    if(flag)
    cout<<"Yes"<<endl;
    else
     cout<<"No"<<endl;
     }
    return 0;
}