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)

//in the name of GOD//

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie();
#define sz(s) s.size()

int main(){
  int a,b,ans = 0;
  bool f;
  cin >> a >> b;
  for(int i = 1;i <= min(a,b); i++){
    if(a % i == 0 && b % i == 0){
      f = true;
      for(int j = 2;j < i;j++){
        if(i % j == 0){
          f = false;
        }
      }
      if(f)
        ans++;
    }
  }
  cout << ans;
}