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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

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

int64_t gcd(int64_t a, int64_t b) {
    if (a < b) {
        return gcd(b, a);
    }
    if (a % b == 0) {
        return b;
    } else {
        return gcd(b, a % b);
    }
}

int main() {
    int64_t a, b;
    cin >> a >> b;
    int64_t c = gcd(a, b);
    int64_t rc = sqrt(c);
    int64_t ans = 1;
    int64_t i = 2;
    while (c > 1) {
        if (c % i == 0) {
            while (c % i == 0) {
                c /= i;
            }
            ans++;
        }
        i++;
    }
    cout << ans << endl;
}