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>
using namespace std;
long int gcd(long int a, long int b) {
  if(a < b) return gcd(b, a);
  long int r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}
int main() {
    long int A, B;
    cin >> A >> B;
    if(gcd(A,B) == 1) {
        cout << 1 << endl;
        return 0;
    }
    int countv = 1;
    for(long int i = 2;i <= max(A,B);i++) {
        if(A % i == 0 && B % i == 0) {
            A /= i;
            B /= i;
            countv++;
            if(gcd(A,B) == 1) {
                cout << 1 << endl;
                return 0;
            }
        }
    }
    cout << countv << endl;
}