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: Runtime Error)

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

int main(){
  long long a, b; cin >> a >> b;
  if(a < b){
    long long tmp = b;
    b = a; a = tmp;
  }
  while(a % b != 0){
    long long u = a % b;
    a = b; b = u;
  }
  vector<int> s(sqrt(b) + 2, 1);
  s.at(0) = 0; s.at(1) = 0;
  for(long long i = 0; i < s.size(); i++){
    if(s.at(i) == 1){
      long long j = 2 * i;
      while(j < s.size()){
        s.at(j) = 0;
        j += i;
      }
    }
  }
  long long out = 0;
  for(int i = 0; i < sqrt(b) + 2; i++){
    if(s.at(i) == 1){
      if(b % i == 0) out++;
    }
  }
  if(out == 0){
    cout << 2;
    return 0;
  }
  cout << out + 1;
}