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: Accepted)

#include <bits/stdc++.h>

using namespace std;

set<long long> s, com;

int main() {
  long long a, b;
  cin >> a >> b;
  long long x = sqrt(a);
  for (long long i = 1; i <= x; i++) {
    if (a % i == 0) {
      s.insert(i);
      if (a / i != i) s.insert(a / i);
    }
  }
  x = sqrt(b);
  for (long long i = 1; i <= x; i++) {
    if (b % i == 0) {
      if (s.find(i) != s.end()) com.insert(i);
      if (b / i != i) if (s.find(b / i) != s.end()) com.insert(b / i);
    }
  }
  for (auto i : com) {
    long long rem = -1;
    for (auto j : com) {
      if (i == j) continue;
      long long g = __gcd(i, j);
      if (g != 1) {
        rem = j;
        break;
      }
    }
    if (rem != -1) com.erase(rem);
  }
  cout << com.size() << '\n';
  return 0;
}