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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, a, b) for (int i = (a); i < (b); ++i)

using namespace std;
using ll = long long;

ll gcd (ll x, ll y) {
  if (x % y == 0) return y;
  return gcd(y, x % y);
}

bool is_prime (ll n) {
  if (n == 1) return false;
  if (n == 2) return true;

  ll sqrtn = (ll) ceil(sqrt(n));

  for (ll i = 2; i <= sqrtn; i++) {
    if (n % i == 0) {
      return false;
    }
  }

  return true;
}

/**
 * main
 */
int main () {
  ll A; cin >> A;
  ll B; cin >> B;
  ll g = gcd(A, B);
  ll sqrtg = (ll) ceil(sqrt(g));
  ll cnt = 0;

  for (ll p = 1; p <= sqrtg; p++) {
    if (!(is_prime(p) || p == 1)) continue;
    if (g % p == 0) {
      ll q = g / p;
      
#ifdef DEBUG
      printf("%lld\n", p);
#endif
      cnt++;
      if (q > sqrtg && is_prime(q)) cnt++;
    }
  }

  cout << cnt << endl;
}