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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>

using namespace std;




int GCD(int a, int b){
    int temp;

    //a<bとする
    if(a<b) return GCD(b,a);
    if(b>0) return( GCD(b, a%b) );
    else    return a;
}

int numberOfDSCD(int a){
    int num=1;
    int fa = a;

    if(a=1){
        return a;
    }
    for(int i=2; i<=fa/2; i++){
        if(a%i==0){
            num++;
            while(a%i == 0){
                a/=i;
            }
        }
    }
    return num;
}


int main(){
    int A,  B;

    //入力
    cin >> A >> B;


    //処理
    cout << numberOfDSCD( GCD(A, B) )<< endl;


    return 0;
}