CodeForceS#276-A

A. 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".

Sample test(s)

input

1 5

output

No

input

3 6

output

Yes

开始做时，一直没理解那个取模是怎么回事，

 #include <iostream>

 using namespace std;

 int main()

 {

     long long a, m;

     bool flag = false;

     while(cin >> a >> m)

     {

         for(int i=; i<=; i++, a = a + (a%m)) ///a的这句话就是题意

             if(a % m == )

                 flag = ;

         cout << (flag ? "Yes" : "No") << endl;

     }

     return ;

 }

巴特西

CodeForceS#276-A

最新文章

热门文章