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>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repi(i, s, n) for (int i = (s); i < (n); i++)
#ifdef LOCAL
#define INPUT_FILE                              \
  ifstream in("atcoder-problems/abc142_d.txt"); \
  cin.rdbuf(in.rdbuf());
#define print_vec(v)                        \
  rep(l, v.size()) { cout << v[l] << " "; } \
  cout << endl;
#else
#define INPUT_FILE
#define print_vec(v)
#endif
#define CIN_OPTIMIZE \
  cin.tie(0);        \
  ios::sync_with_stdio(false);
typedef pair<int, int> P;
typedef long long ll;
typedef pair<ll, ll> pl;
const int INF = 100100100;
const ll LINF = 1e18 + 100;
const int MOD = 1e9 + 7;

bool IsPrime(ll num) {
  if (num < 2)
    return false;
  else if (num == 2)
    return true;
  else if (num % 2 == 0)
    return false;  // 偶数はあらかじめ除く

  ll sqrtNum = ceil(sqrt(num));
  for (ll i = 3; i <= sqrtNum; i += 2) {
    if (num % i == 0) {
      // 素数ではない
      return false;
    }
  }

  // 素数である
  return true;
}

int main() {
  INPUT_FILE CIN_OPTIMIZE;

  ll a, b;
  cin >> a >> b;

  ll ans = 1;
  ll n = min(a, b);
  ll sqrn = ceil(sqrt(n));
  repi(i, 2, sqrn + 1) {
    if (a % i == 0 && b % i == 0 && IsPrime(i)) ans++;
  }
  ll m = max(a, b);
  if (m % n == 0) ans++;

  cout << ans << endl;
}