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: Time Limit Exceeded)

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

int main() {
  //素数の公約数の数+1
  long long A,B;
  cin >> A >> B;
  vector<long> s(12,0);
  int c =0;
  for(int i = 2;i<A+1;i++)
  {
    while(A % i == 0)
    {
      A /= i;
      s.at(c) = i;
    }
    if(s.at(c) == i)
    {
      c++;
    }
  }
  vector<long> s2(12,0);
  int c2 =0;
  for(int i = 2;i<B+1;i++)
  {
    while(B % i == 0)
    {
      B /= i;
      s2.at(c2) = i;
    }
    if(s2.at(c2) == i)
    {
      c2++;
    }
  }
  int cnt = 1;
  for(int i = 0 ; i < 12 ; i++)
  {
    if(s.at(i) == 0)
    {
      cout << cnt << endl;  
      return 0;
    }
    for(int j = 0;j<12;j++)
    {
       if(s.at(i) == s2.at(j))
       {
         cnt++;
       }
    }
  }
  cout << cnt << endl;  

}