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: Wrong Answer)

#include <iostream>
#include <algorithm>

using namespace std;

int main(void){
    ios::sync_with_stdio(false);
    cin.tie(0);

    int a, b;
    cin >> a >> b;

    //a<b
    if(b<a) swap(a,b);

    int gcd = a;

    while(a!=0){
        int mod=b%a;
        if(mod==0)
            gcd = a;
        b=a;
        a=mod;
    }

    int cnt = 0;

    for(int x=2; x<=gcd; x++){
        if(gcd%x == 0){
            cnt++;

            while(gcd % x == 0)
                gcd /= x;
        }
    }

    if(gcd >= 1)
        cnt++;

    cout << cnt << endl;

    return 0;
}