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;
signed main(){
	int a,b,z,ans=0;
  	cin>>a>>b;
  	z=a%b;
  	while(z!=0){
      a=b;
      b=z;
      z=a%b;
    }
  int putch[(b+1)];
  for(int i=1;i<=b;i++){
    putch[i]=1;
    if(b%i!=0)
      putch[i]=0;
  }
	for(int i=2;i<=b;i++){
      	for(int j=(i+i);j<=b;j+=i)
          putch[j]=0;
    }	
    cout<<endl;
  for(int i=1;i<=b;i++)
   	 ans+=putch[i];
  
	cout<<ans<<endl;
	  return(0);
}