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>

using namespace std;

int main()
{
    long long int a, b;
    cin >> a >> b;
    long long int c = a;
    if(a > b){
        c = b / 2;
    }
    
    std::vector<long long int> v;
    v.push_back(1);
    std::vector<long long int>::iterator itr;
    for(int i=2; i<=c; i++){
        if(a%i == 0 && b%i == 0){
            //cout << "------" << endl;

            bool ok = true;
            itr = v.begin();
            while(itr != v.end()){
                //cout << i << ":" << *itr << endl;
                if(*itr == 1){
                    itr++;
                    continue;
                }
                if(i%(*itr) == 0){
                    ok = false;
                    break;
                }
                ++itr;
            }
            if(ok == true){
                //cout << i << endl;
                v.push_back(i);
            }
        }
    }
    //cout << endl;
    cout << v.size() << endl;
    
    return 0;
}