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 <iostream>
#include <vector>
int A;
int B;
int gcd_ans;
std::vector<int> ans;

int gcd(int a,int b)
{
	if (a%b==0)
	{
		return(b);
    }
	else
	{
		return(gcd(b,a%b));
    }
}

int su(int a)
{
    for(int i = 3; i < (a/2); i++){
        if( (a % i ) != 0 )
            return 0; //素じゃなし
    }
    return 1; //素
}

int main(void){
    std::cin >> A >> B;;

    gcd_ans = gcd(A,B);
    //std::cout << "GCD:" << gcd_ans << std::endl; 

    ans.push_back(1);
    ans.push_back(2);
    //std::cout << 1 << " " << 2 << " ";

    for(int i = 3; i <= gcd_ans; i++){
        //奇数か偶数か？
        if( i % 2 != 0){
            //約数か？
            if (gcd_ans % i == 0){
                //素か？
                if ( su(i) ){
                    ans.push_back(i);
                    //std::cout << i << " ";
                } else {
                    //std::cout <<  "素じゃない ";
                }
            } else {
                //std::cout << "約数じゃない ";
            }
            
    
        } else {
            //std::cout <<  "偶数だ ";
        }
    }
    
    std::cout << ans.size();
 
    return 0;
}