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;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;
 
template<class Z> map<Z, int> factorize(Z n) {
  map<Z, int> res;
  for (Z i = 2; i * i <= n; ++i) while (n % i == 0) ++res[i], n /= i;
  if (n != 1) ++res[n];
  return res;
}
 
int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  lint a, b; cin >> a >> b;
  auto mp = factorize(__gcd(a, b));
  cout << mp.size() + 1 << '\n';
}