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>

#define REPI(i, n) for(int i = 0; i < n; i++)
#define REPJ(j, n) for(int j = 0; j < n; j++)

using ll = long long;
using namespace std;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll a, b; cin >> a >> b;
    ll small = min(a, b);
    vector<ll> vc;
    bool flag;

    for(int i = 1; i <= small; i++){
        if((a % i | b % i) == 0){
            if(i % 2 != 0){
                vc.push_back(i);
            }else if(i == 2) vc.push_back(i);
        }
    }

    int vc_length = vc.size();
    REPI(i, vc_length) cout << vc[i] << endl;
    int negative = 0;
    for(int i = 1; i < vc_length; i++){
        for(int j = i; j < vc_length; j++){
            if(i == j || vc[i] == 0) continue;
            if(vc[j] % vc[i] == 0 && vc[j] != 0){
                vc.erase(vc.begin() + j);
                vc_length = vc.size();
            }
        }
    }

    cout << vc_length;
}