Factory

One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were x details in the factory storage, then by the end of the day the factory has to produce (remainder after dividing x by m) more details. Unfortunately, no customer has ever bought any mythical detail, so all the details produced stay on the factory.

The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by m).

Given the number of details a on the first day and number m check if the production stops at some moment.

Input

The first line contains two integers a and m (1 ≤ a, m ≤ 10⁵).

Output

Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No".

Examples

Input

1 5

Output

No

Input

3 6

Output

Yes

此题就是给你a，m。每天进行a = a+（a%m）操作。让你判断，如果最终a不能被m整除，输出No，否则输出yes。
代码如下

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
bool vis[100010];
int a,m;
int main()
{
    cin>>a>>m;
    memset(vis,0,sizeof(vis));
    if(a%m==0)
    {
        cout<<"Yes"<<endl;
        return 0;
    }
    while(1)
    {
        a=(a+a%m)%m;
        if(a==0)
        {
            cout<<"Yes"<<endl;
             return 0;
        }
        if(vis[a])
        {
            cout<<"No"<<endl;
            return 0;
        }
        vis[a]=1;
    }
    return 0;   
}