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: Runtime Error)

                                                    #include <iostream>
#include <iomanip>
#include <bits/stdc++.h>
#include <vector>
#include <iterator>
#include <string>
#define pan(i,n) for(int i=0;i<n;i++)
#define ll long long int
#define vi vector<int>
#define pb push_back
using namespace std;
 
ll gcd(ll a , ll b){
    if(b==0)
        return a;
    else {
        return gcd(b,a%b);
    }
}


int main(){
    ll a,b;
    cin>>a>>b;
    ll n=gcd(a,b);
    
    bool prime[n+1];
    memset(prime,true,sizeof(prime));
    ll count=1;
    for (int p=2; p*p<=n; p++) 
    { 
        if (prime[p] == true) 
        { 
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    }
    for(ll i=2;i<=n;i++){
        if(prime[i]==true && n%i==0){
            count++;
        }
    }
    cout<<count;
}