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 <bits/stdc++.h>

using namespace std;
using ll = long long;

int main()
{
    ll i;
    ll A,B;
    cin >> A >> B;

    ll primecnt=0;
    ll max=B;
    vector<ll> prime;
    if(A>B) max=A;

    for(i=1;i<=max;i++)
    {
        if(A % i == 0 && B % i == 0)
        {
            if(i < 2)
            {

            }
            else if(i == 2)
            {
                prime.push_back(i);
            }
            else if(i%2!=0)
            {
                ll j;
                bool flag=true;
                for(j=0;j<prime.size();j++)
                {
                    if(i % prime[j] == 0)
                    {
                        flag=false;
                        break;
                    }
                }

                if(flag)
                {
                    prime.push_back(i);
                }
            }
        }
    }

    
    cout << prime.size() + 1 << endl;

    return 0;
}