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>
using namespace std;
int main (){
  long long A,B;
  cin>>A>>B;
  long long AA, BB;
  AA=A,BB=B;
  long long k=1;
  long long j=0;
  while(k>0){
      j=A/B;
      k=A%B;
      A=B;
      B=k;
  }

  int count=1;
  long long E=A;
  int x=max(sqrt(AA),sqrt(BB));
  //cout<<x;
    for(int i=2;i<x+1;i++){
          if(E%i==0){
              while(E%i==0){
                  E=E/i;
              }
              count++;

          }
        if(E<i){
            break;
        }
        
      }
  cout<<count<<endl;
  
}