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)

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


ll gcd(ll a, ll b) {
  return b != 0 ? gcd(b, a % b) : a;
}

int main(void)
{
    ll A, B;
    cin >> A >> B;
    ll C = gcd(A, B);
    ll n = C;
    //cout << C << endl;
    int count = 0;
    for(ll i = 2; i * i <= n; i++)
    {
        if(C % i == 0)
        {
            count++;
            while(C % i == 0)
            {
                C /= i;
            }
        //cout << i << " " << C << endl;
        }
    }
    
    cout << count + 1 << endl;
}