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: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

vector<long> soinsu(long a);

int main(){
    long A, B;

    cin >> A >> B;
    
    vector<long> al, bl;

    al = soinsu(A);
    bl = soinsu(B);

    long count = 0;
    for(auto it = al.begin(); it != al.end(); ++it){
        if(find(bl.begin(), bl.end(),*it) != bl.end()){
            count ++;
        }
    }

    cout << count << endl;

    return 0;

}

vector<long> soinsu(long a){
    long i = 2;
    vector<long> l;
    l.push_back(1);

    while(a != 1){
        if(a % i == 0) l.push_back(i);
        while(a % i == 0){
            a /= i;
        }
        i++;
    }

    return l;

}