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(){
  long A,B,AbMin,ans_total=1;
  cin >> A >> B;
  vector<long> ans;
  AbMin=min(A,B);
  for(long i=1;i<=AbMin;i++){
    if(A%i==0 && B%i==0){
      ans.push_back(i);
    }
  } 
  int cnt;
  for(int i=1;i<ans.size();i++){
    cnt=0;
    for(int ii=1;ii<i+1;ii++){
      if(ans.at(i)%ans.at(ii)!=0){
      }else{
        cnt++;
      }
    }
    if(cnt==1) ans_total++;
  }
  cout << ans_total << endl; 
}