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>
using namespace std;

// vector<int> finder(vector<int> vec, int number) {
//   vector<int> ans;
//   for(int i=0;i<vec.size();i++){
//     if(vec[i]%number==0){
//       ans.push_back(i);
//     }
//   }
//   return ans;
// }


int main() {
  int a,b;
  cin>>a>>b;
  bool an=false;
  vector<int> pol;
  for(int i=1;i<=min(a,b);i++){
    if(a%i==0&&b%i==0){
      for(int a=2;a<i;a++){
        if(i%a==0) {
          an=true;
          break;
        }
      }
      if(an){
      }
      else{
        pol.push_back(i);
      }
    }
  }
  // vector<int> bat;
  // for(int i=0;i<pol.size();i++){
  //   for(int j=i+1;j<pol.size();j++){
  //     for(int x=2;x<=min(pol[i],pol[j]);x++){
  //       if(pol[i]%x==0&&pol[j]%x==0){
  //         bat.push_back(i);
  //         bat.push_back(j);
  //       }
  //     }
  //   }
  // }
  // sort(bat.begin(),bat.end());
  // for(int i=0;i<bat.size();i++){
  //   pol.erase(pol.begin()+bat[i]);
  // }
  cout<<pol.size()+1;
}