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>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
#include <functional>
using namespace __gnu_pbds; 
using namespace std;



typedef tree<long long, null_type, less<long long>, rb_tree_tag, 
             tree_order_statistics_node_update> 
    new_data_set; 
 
 
//defines
typedef long long ll;
typedef long double ld;

ll a,b,c,n,m,k,t=1;

void solve()
{
 cin>>n>>m;
 if(n>m)swap(n,m);
 c=1;
 if(n%2==0 and m%2==0)c++;
 if(n%3==0 and m%3==0)c++;
 
 for(ll i=5;i<=__gcd(n,m);i+=6)
 {
     if(n%i==0 and m%i==0)c++;
     
     if((i+2)<=__gcd(n,m) and n%(i+2)==0 and m%(i+2)==0)c++;
 }
 cout<<c;
 if(t>0)cout<<"\n";
}

int32_t main()
{
#ifdef ONPC
    freopen("input.txt", "r", stdin);
#endif
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

  //cin>>t;
    while(t--)
    {
        solve();
    }
  //cout<<sum;
        
    return 0;
}