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>

#define REP(i, n) for (int i = 0; i < n; ++i)

#define REPR(i, n) for (int i = n; i >= 0; --i)

#define _REP(i, m, n) for (int i = m; i < n; ++i)

#define INF 2e9

using namespace std;
long long  e_gcd(long long  a, long long  b)
{
    if (a < b)
        return e_gcd(b, a);
    long long  r;
    while ((r = a % b))
    {
        a = b;
        b = r;
    }
    return b;
}

int main()
{
    long long a, b;
    cin >> a >> b;

    int num = e_gcd(a, b);

    long long ans = 1;
    for (long long i = 2; i * i <= num; i++)
    {
        if (num % i == 0)
        {
            while (num % i == 0)
            {
                num /= i;
            }
            ans++;
        }
    }

    if (num != 1)
        ans++;
    //  REP(i,cnt)cout << g[i] << " ";
    //  cout << endl;

    cout << ans << endl;

    return 0;
}