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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)

long long gcd(long long a, long long b){
    
    if (a < b)
    {
        long long tmp = a;
        a = b;
        b = tmp;
    }

    return b ? gcd(b, a % b) : a;
}


int main(){    
    long long A,B;
    cin >>A >>B;
    long long a = gcd(A,B);
    long long b=2,c=0;
    bool f = false;
    while(a%b ==0){
        a/=b;
        f = true;
    }
    b++;
    if(f ==true){
        c++,f= false;
    }
    while(a!=1){
        while(a%b ==0){
            a/=b;
            f = true;
        }
        b+=2;
        if(f ==true){
            c++,f= false;
        }
    }
    cout << c+1;
}