Vanya and Scales(思维)

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.

题解：很棒的一个题，初一看完全没思路，到最后也没搞定。。。这个题的意思是给w的（0-100）次方，0-100次方只用一次，使天平平衡，

转化一下就是一个w进制的数能否通过加减w的（0-100）次方，0-100次方只用一次，使之等于0；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<algorithm>
using namespace std;
int a[110];
int main(){
int w, m;
while(~scanf("%d%d", &w, &m)){
int k = 0;
while(m){
a[k++] = m % w;
m /= w;
}
int ans = 1;
for(int i = 0; i < k; i++){
if(a[i] == w){
a[i] = 0;
a[i + 1]++;
continue;
}
if(a[i] <= 1){
continue;
}
else if(a[i] == w - 1){
a[i + 1]++;
a[i] = 0;
continue;
}
else
ans = 0;
}
if(ans)
puts("YES");
else
puts("NO");
}
return 0;
}