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 mp make_pair
#define pb push_back
#define rep(i,n) for(int i=0;i<(n);i++)
typedef long long int ll;
using namespace std;

//ユークリッドの互除法
ll gcd(int x,int y){
    if(y==0) return x;
    return gcd(y,x%y);
}

int main(){
cin.tie(0);
ios::sync_with_stdio(false);
ll A,B;
cin>>A>>B;
ll k=gcd(A,B);
vector<ll> cd;
cd.pb(1);
for(ll i=2;i*i<=k;i++){
    if(k%i==0){
    cd.pb(i);
    while(k%i==0){
        k/=i;
    }
    }
}
if(k>1)cd.pb(k);
cout<<cd.size()<<endl;
 
//system("pause");
return 0;
}