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;

int64_t gcd(int64_t a, int64_t b) {
  auto x = minmax(a, b);
  auto lesser = x.first;
  auto greater = x.second;
  if (greater % lesser == 0)
    return lesser;
  return gcd(greater - lesser, lesser);
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);

  int64_t N, M;
  cin >> N >> M;

  auto g = gcd(N, M);
  int count = 1;

  if (g % 2 == 0) {
    ++count;
    for (; g % 2 == 0; g /=2) {
    }
  }


  auto limit = sqrt(g + 1);

  int64_t num = 3;
  for (; g != 1; num += 2) {
    if (num > limit) {
      ++count;
      break;
    }
    // cout << " g " << g << endl;
    if (g % num == 0) {
      // cout << " num " << num << endl;
      ++count;
      g /= num;
      for (; g % num == 0; g /= num) {
      }
      limit = sqrt(g + 1);
    }
  }

  cout << count << endl;
}