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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for(int i = 0; i < n; i++)
#define MOD 1000000007
typedef long long ll;

vector<int> calc(int n){
    int d = 2;
    vector<int> p;
    bool flag = false;
    while(n >= d*d){
        if(n % d == 0){
            n /= d;
            if(!flag){
                p.push_back(d);
                flag = true;
            }
        }else {
            d++;
            flag = false;
        }
    }
    p.push_back(n);
    return p;
}
int main() {
    int a,b,ans = 0;
    cin >> a >> b;
    vector<int> av = calc(a);
    vector<int> bv = calc(b);
    rep(i, av.size()){
        rep(j, bv.size()){
            if(av[i] == bv[j]){
                ans++;
                break;
            }
        }
    }
    cout << ans+1 << endl;
}