CF 552-C. Vanya and Scales

http://codeforces.com/problemset/problem/552/C

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 than2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with massm 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 masses9 and 1, correspondingly. Then7 + 3 = 9 + 1.

Note to the second sample test. One pan of the scales can have an item of mass99 and the weight of mass 1, and the second pan can have the weight of mass100.

Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.

一個天平，100個weights，重量爲w的0次冪到100次冪 各一個， 通過天平和一些weights，問測出重量爲m的物體是否可行。

方法： 將m轉化成w進制的數。 由於每種砝碼只有1個。所以各個位如果 是0，代表該砝碼沒用到， 1代表砝碼與物體異側，w-1代表砝碼與物體同側。其他情況則不能構成m

#include<bits/stdc++.h>
#define For(i,a,b) for(int (i)=(a);(i) < (b);(i)++)
#define rof(i,a,b) for(int (i)=(a);(i) > (b);(i)--)
#define IOS ios::sync_with_stdio(false)
#define lson l,m,rt <<1
#define rson m+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

const int maxn = 2e2+10;
const int INF =0x3f3f3f3f;
ll w,m;
int main()
{
    cin>>w>>m;
    ll bit[maxn];
    mem(bit,0);
    int len=0;
    while(m){
        bit[len++]=m%w;
        m/=w;
    }
    For(i,0,len){
        if(bit[i]==0||bit[i]==1) continue;
        if(bit[i]>=w-1) {
            bit[i+1]++;len++;
            continue;
        }
        else{
            cout<<"NO"<<endl;
            return 0;
        }
    }
    cout<<"YES"<<endl;
    return 0;
}