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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

template <typename T> T gcd(T a, T b) {
  if (a < b) return gcd(b, a);
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll A, B;
  cin >> A >> B;
  ll gcdn = gcd(A, B);
  ll gcdcopy = gcdn;
  int res = 1;
  for (int i = 2; i * i <= gcdn; i++) {
    if (gcdcopy % i != 0) continue;
    res++;
    while (gcdcopy % i == 0) {
      gcdcopy /= i;
    }
  }
  if (gcdcopy > 1) {
    res++;
  }
  cout << res << '\n';
  return 0;
}