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: Wrong Answer)

// CPP program to illustrate 
// gcd function of C++ STL 
#include <iostream> 
#include <algorithm>
#include <cmath>
#include <vector>  
#define ll long long
  
using namespace std; 

int main() {
    ll A,B;
    cin >> A >> B;
    ll D = __gcd(A,B);
    double mid = sqrt(D);
    ll count = 0;
    // cout << D << endl;
    if (D > 1){
        for (ll x = 2; x <= mid+1; x++) {
            // cout << x << ' ' << D << endl;
            if (D % x == 0) {
                count++;
                while(D % x == 0){
                    D /= x;
                }
                
            }
        }
    }

    cout << count + 1 << endl;

}