Problem:
Score : 400 points

Problem Statement

Given are positive integers A and B.

Let us choose some number of positive common divisors of A and B.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisor

An integer d is said to be a common divisor of integers x and y when d divides both x and y.

Definition of being coprime

Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.

Definition of dividing

An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.

Constraints

All values in input are integers.

1 \leq A, B \leq 10^{12}

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

Sample Input 1

12 18

Sample Output 1

3

12 and 18 have the following positive common divisors: 1, 2, 3, and 6.

1 and 2 are coprime, 2 and 3 are coprime, and 3 and 1 are coprime, so we can choose 1, 2, and 3, which achieve the maximum result.

Sample Input 2

420 660

Sample Output 2

4

Sample Input 3

1 2019

Sample Output 3

1

1 and 2019 have no positive common divisors other than 1.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
int gcd(int a , int b){
 int r, tmp;
 if(a<b){
    tmp = a;
    a = b;
    b = tmp;
  }
  r = a % b;
  while(r!=0){
    a = b;
    b = r;
    r = a % b;
  }

  return b;
}

//int v,w; cin>>v>>w;
int main()
{
  int v,w,k=0; cin>>v>>w;
	char  c[max(v,w)];
    int   n0,n;   // 整数
    int   ns=2;   // 素因数
    int   j;      // 乗数

   n0=gcd(v,w);

    for( n=n0,ns=2; n>=ns; ns++ )
      {
    //  while(n%ns==0)n/ns;
        for(; n%ns==0;) {           // 割り切れる間繰り返す
           n/=ns;
        }

      //  if ( j==0 ) continue;               // 1回も割り切れなかった

        //printf("素因数:%d  乗数:%d\n",ns,j);
      k++;
     }
  cout<<k+1;
}