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 int A,B;
  cin >> A >> B;


  vector<long int> O;

  O.push_back(1);
  if ( (A > 1) & (A%2 == 0) ) O.push_back(2);
  
  int k = 0;
  if ( A > 2 )
    for (long int i = 3; i <= sqrt(A)+1; i=i+2 ) {
      if ( A%i == 0 ) {
        for (long int j = 1; j < O.size(); ++j ) {
          if ( i%O[j] == 0 ) {
             k = 1;
             break;
          }
          if ( k == 0 ) O.push_back(i); 
          k = 0;
        }
      } 
    }
  
  int C = 1;
  for (long int i = 1; i < O.size(); ++i )
    if ( B%O[i] == 0 ) ++C;

  cout << C;
}