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 <iostream>
#include <vector>
using namespace std;

int gcd(int a, int b){
  int t;
  while(b!=0){
    t = a%b;
    a = b;
    b = t;
  }
  return a;
}

int main(){
  int A,B;
  cin >> A >> B;
  vector<int> kouyaku;
  if(A<B){
    for(int i=1; i<=A; i++){
      if(A%i==0 && B%i==0){
        kouyaku.emplace_back(i);
      }
    }
  }else{
    for(int i=1; i<=B; i++){
      if(A%i==0 && B%i==0){
        kouyaku.emplace_back(i);
      }
    }
  }
  
  
  int N = kouyaku.size();
  for(int j=1; j<N; j++){
    cout << kouyaku[j] << ": ";
    for(int k=j+1; k<N; k++){
      if(gcd(kouyaku[j],kouyaku[k])==1){
		cout << kouyaku[k] << ",";
      }
    }
    cout << endl;
  }
  
  
  return 0;
}