Codeforces Round #308 (Div. 2)----C. Vanya and Scales

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 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 test(s)

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.

开始是没有读懂题，好奇怪，为什么老是这样子，后来才看明白，题意大概是这样子的：

现在有一个重量为m的物品 放到天平上 利用一些砝码使得天平平衡砝码的重量为w的i次方 问有没有一种方式使得天平平衡 。

然后是做法，额，是个多项式的问题，每个砝码只有三种选择 不放 放左边 放右边  

其实就是等式a0*w^0+a1*w^1+……+an*w^n=m有没有解的情况 ai的值只能是0 1 -1 

对于这个等式 让m不断的除以w 则模m得到的值只能是 0 1 w-1(-1) 

#include<string.h>
#include<stdio.h>
#define LL __int64
LL w,m;
int main()
{
    scanf("%I64d%I64d",&w,&m);
    if(w<=3)
    {
        printf("YES
");
        return 0;
    }
    while(m)
    {
        if(!((m-1)%w)) m--;
        else if(!((m+1)%w)) m++;
        else if(m%w) {printf("NO
");return 0;}
        m=m/w;
    }
    printf("YES
");
    return 0;
}