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 <cstdio>
#include <iostream>
#include <cmath>
#define rep(i,n,m) for(ll i = (n); i<(m); i++)
using ll = long long int;

int main(void){
  ll n, k;
  std::cin >> n >> k;
  if (n>k){
    ll tmp = k;
    k = n;
    n = tmp;
  }
  if (n < 3){
    std::cout << n << std::endl;
    return 0;
  }
  ll N = std::sqrt(k)+1;
  ll* array = new ll[N]();
  rep(i,0,N)
    array[i] = -1;
  array[0] = 1;
  ll count = 1;
  if (n%2 == 0 && k%2 == 0){
    array[1] = 2;
    count++;
  }
  ll ans = 2;
  ll yaku[3];
  for ( ll i = 3; i <= N; i+=2){
    if (n%i == 0 && k%i == 0){
      ll lim = 1;
      yaku[0] = i;
      if (k%(n/i) == 0){
        yaku[lim] = n/i;
        lim++;
      }
      if (n%(k/i) == 0){
        yaku[lim] = n/k;
        lim++;
      }
      rep(d,0,lim){
        bool can = true;
        for (ll j = 1;array[j] != -1; j++){
          if (yaku[d]%array[j] == 0){
            can = false;
            break;
          }
        }
        if (can){
          array[count] = yaku[d];
          count++;
        }
      }
    }
  }
  for(ans = 0; array[ans] != -1; ans++);
  std::cout << ans << std::endl;
  return 0;

}