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() {
  int64_t A,B;
  cin >> A >> B;
  
  if(A>B)swap(A,B);
  
  vector<int64_t> V{};
  V.push_back(1);
  if(A!=1)V.push_back(A);
  
  for(int i=2;i<A;i++)
  {
    if(i<1){cout << "a" << endl;return 0;}
    if(A%i==0)
    {
      V.push_back(i);
      if(i!=A/i)V.push_back(A/i);
    }
    if(i*i>A)break;
  }
  
  for(int64_t i=1;i<V.size();i++)
  {
    if(V.at(i)<1){cout << i<<"b" << V.at(i) << endl;return 0;}
    if(B%V.at(i)!=0)
    {
      V.erase(V.begin()+i);i--;
    }
  }
  
  sort(V.begin(),V.end());
  
  for(int64_t i=1;i<V.size();i++)
  {
    for(int64_t j=i+1;j<V.size();j++)
    {
      if(V.at(i)<1){cout << "c" << endl;return 0;}
      if(V.at(j)%V.at(i)==0)
      {
        V.erase(V.begin()+j);
        j--;
      }
    }
  }
  
  cout << V.size() <<endl;
  
  return 0;
}