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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;

int64_t gcm(int64_t x,int64_t y)
{
  if(x%y==0)
   return y;
  return gcm(y,x%y);
}

int main()
{
  int64_t a,b,c,d,ans;
  ans=1;
  cin>>a>>b;
  c=gcm(a,b);
  d=c;
  if(c!=1)
  {
    for(int64_t i=2;i<=sqrt(d);i++)
    {
      if(c%i==0)
      {
        ans++;
        while(c%i==0)
        {
          c=c/i;
        }
      }
    }
    if(c==gcm(a,b))
      ans++;
  }
  cout<<ans<<endl;
}