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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
   ll a,b,c,cnt=0;
   cin >> a >> b;
   c = __gcd(a,b);
   ll d[c];
   for(int i=0;i<c;i++){
        d[i]=0;
   }
   for(ll k=2;k<=c;k++){
       for(ll i=1;i<=c;i++){
           if(i%k==0) d[i-1]++;
       }
   }
   ll e = c;
   set<ll> f;
   for(int i=0;i<c;i++){
       if(d[i]==1){
           while(e%(i+1)==0){ 
               e/= (i+1);
               f.insert(i+1);
           }
       }
   }
   cout << f.size() +1 << endl;
}