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>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
int INF = 1000000007;
 
int main(){
  cin.tie(nullptr);
  ios::sync_with_stdio(false); 
 
  int64_t N, M;
  cin >> N >> M;
  int64_t D = __gcd(N, M);
  int64_t ans = 1;
  for(int i=2;i<= 10000000;i++){
    bool ans1 = true;
    if(D % i == 0){
      for(int j=2;j<= (int)sqrt(i)+100;j++){
        if(j == i) break;
        if(i % j == 0){
          ans1 = false;
          break;
        }
      }
      if(ans1) ans++;
    }
  }
  bool ans1 = true;
  if(D > 10000000){
    for(int64_t j=2;j<= (int64_t)sqrt(D)+100;j++){
        if(j == D) break;
        if(D % j == 0){
          ans1 = false;
          break;
        }
      }
      if(ans1) ans++;
  }
  cout << ans << endl;
}