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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

#define ll long long

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

int main()
{
    ll a, b, n;
    cin >> a >> b;
    n = gcd(a, b);

    ll ans = 1;
    for (ll i = 2; i * i <= n; i++)
    {
        if (n % i == 0) //!(n % i)
        {
            // cout << i << "\n";
            ans++;
            while (n % i == 0) n /= i;
        }
    }

    if (n != 1) ans++;
    cout << ans << "\n";
}