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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i,n) for(int i=0;i<n;i++)

int A, B;
set<int> SA, SB;

signed main()
{
    cin >> A >> B;

    SA.insert(1);
    SB.insert(1);

    for(int i = 2; i <= 1000000; i++)
    {
        while (A % i == 0)
        {
            SA.insert(i);
            A /= i;
        }
        
        while(B % i == 0)
        {
            SB.insert(i);
            B /= i;
        }
    }
    SA.insert(A);
    SB.insert(B);

    vector<int> v;
    set_intersection(SA.begin(), SA.end(), SB.begin(), SB.end(), back_inserter(v));

    cout << v.size() << endl;

    return 0;
}