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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i=0; i<(n); ++i)
typedef long long ll;

ll gcd(ll x, ll y) {return y?gcd(y,x%y):x;}

vector<pair<ll,int>> factorize(ll n){
  vector<pair<ll,int>> res;
  for(ll i=2; i+1 <= n; i++){
    if(n%i) continue;
    res.emplace_back(i,0);
    while(n%i == 0){
      n /= i;
      res.back().second++;
    }
  }
  if(n != 1) res.emplace_back(n,1);
  return res;
}
      


int main() {
  ll a, b; cin >> a >> b;
  ll g = gcd(a, b);
  auto f = factorize(g);
  int ans = f.size() + 1;
  cout << ans << endl;  
  return 0;
  
}