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<algorithm>

using namespace std;
  
vector<int> divisor(int n, int gcd) {
  vector<int> ret;
  for(int i = 1; i * i <= n; i++) {
    if(i > gcd) {
      break;
    }
    if(n % i == 0) {
      ret.push_back(i);
      if(i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}

int main() {
  int A,B;

  cin >> A >> B;

  int gcd = __gcd(A,B);
  
  vector<int> a = divisor(A, gcd);
  vector<int> b = divisor(B, gcd);
  
  vector<int> c;
  set_intersection(a.begin(), a.end()
                     , b.begin(), b.end()
                     , inserter(c, c.end()));

  vector<int> d; // 互いに素な約数

  d.push_back(1);
  
  for(int i=1; i<c.size(); i++){
    if(c.at(i) !=2 && c.at(i)%2 == 0){
      continue;
    }
    bool so = true;

    for(int l=0; l < d.size(); l++){
      if(d.at(l) !=1 && c.at(i)%d.at(l) == 0){
        so = false;
        break;
      }
    }

    if(so) {
      d.push_back(c.at(i));
    }
    
  }
  cout << d.size();
  
 }