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: Accepted)

#include<bits/stdc++.h>
#define QS ios_base::sync_with_stdio(false);
typedef long long ll;
const ll INF = 1e18;
const int MOD = 1e9+7;
using namespace std;

int main(){
//freopen("output.txt", "w", stdout);
    QS
    ll a, b;
    cin >> a >> b;
    ll g = __gcd(a, b);
    ll cnt = 1;
    for(ll k = 2; k * k <= g; ++k){
        if(g%k==0){
            cnt++;
            while(g%k == 0){
                g/=k;
            }
        }
    }
    if(g > 1) cnt++;
    cout << cnt << '\n';
}