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: Accepted)

#include<iostream>

using namespace std;

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

int main(){
    long A, B;
    cin >> A >> B;
    long g = gcd(A, B);
    long temp = g;
    long ans = 0;
    for(long i = 2; i < 10000000; i++){
        if(g % i == 0 ){
            while(g % i == 0){
                g /= i;
            }
            ans++;
        }
    }
    if(g > 10000000 ) cout << ans + 2 << endl;
    else cout << ans+1 <<endl;
    return 0;
}