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

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.

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.

要用质量为 w0,w1,...,w100 的砝码各1个称出重量m，砝码可以放在天平左边也可以放在右边。问是否可以称出，输出YES或NO。

如样例3，7：左边放3和物品，右边放1和9即可。

假设可以称出，则用w进制表示m，每一位上一定是0，1或w - 1，否则一定不行。

而如果某一位是w - 1则说明当前砝码跟物品放在一起，相当于给物品加上了这个砝码的重量。

我们只需要模拟这个过程，提取m的每一位然后计算即可。

转载请注明出处：寻找&星空の孩子 

题目链接：http://codeforces.com/contest/552/problem/C

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