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>
#pragma GCC optimize("Ofast")
using namespace std;

long long euclidean_gcd(long long a, long long b) 
{
  while(1) 
  {
    if(a < b) swap(a, b);
    if(b == 0) break;
    a %= b;
  }
  return a;
}


long long insuu(long long n)
{
	if (n == 1)
  {
    return 1;
  }
	long long ans = 1;
  for (long long i = 2; i * i <= n; ++i)
  {
    if (n % i == 0)
    {
      while (n % i == 0)
      {
        n /= i;
      }
      ans++;
    }
    else
    {
      continue;
    }
    
    if (n == 1LL)
    {
      break;
    }
  }
  return ans;
}

int main()
{
  long long A,B;
  cin >> A >> B;
  long long common =  __gcd(A, B);

  cout << insuu(common) << endl;

}