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>

#define rep(i, n) for (int i = 0; i < (n); i++)
#define repr(i, n) for (int i = (n)-1; i >= 0; i--)

using namespace std;
using ll = long long;
typedef pair<int, int> Pair;

ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

ll cnt = 1;

int main() {
  ll A, B;
  cin >> A >> B;

  ll g = gcd(A, B);
  ll dv = 2;

  while (g > 1) {
    if (g % dv == 0) {
      while (g % dv == 0) {
        g /= dv;
      }
      cnt++;
    }
    dv++;
    if (dv > 1000000) break;
  }

  if (g > 1) cnt++;
  cout << cnt << endl;
  return 0;
}