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<iostream>
#include<sstream>
#include<ios>
#include<iomanip>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<deque>
#include<map>
#include<string>
#include<numeric>
#include<set>
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rep2(i, m, n) for(int i = (int)(m); i < (int)(n); i++)
#define rep_inv(i, n, m) for(int i = (int)(n); i > (int)(m); i--)
using namespace std;
using vi = vector<int>;
using vc = vector<char>;
using vvi = vector<vector<int>>;
using vvc = vector<vector<char>>;
using pii = pair<int, int>;
using vpii = vector<pii>;

int main(){
  long A, B;
  cin >> A >> B;

  long C = min(A, B);

  if(C == 1){
    cout << 1 << endl;
    return 0;
  }

  int len = (int)ceil(sqrt(C));
  int cnt = 1;
  vi primes(len+1, 1);
  bool prime = true;
  
  rep2(i, 2, len+1){
    if(primes[i] == 1){      
      int sweep = 2*i;
      
      while(sweep <= len){
		primes[sweep] = 0;
		sweep += i;
      }

      if(A%i == 0 && B%i == 0){
		prime = false;
		cnt++;
      }
      
    }
  }

  if(prime){
    if(max(A, B) % C == 0)
      cnt++;
  }

  cout << cnt << endl;
  
  return 0;
}