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, n) for (int i = 0; i < (n); ++i)

using namespace std;

using ll = long long;
using P = pair<int, int>;

// 約数を列挙
vector<ll> enumerateDivisors(ll n) {
  vector<ll> res;
  for (ll i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      res.push_back(i);
      // 対応する約数も追加(自分自身でなければ)
      if (n / i != i) {
        res.push_back(n / i);
      }
    }
  }
  // 昇順にソート
  sort(res.begin(), res.end());
  return res;
}

// 素数か判定する
bool isPrime(ll n) {
  if (n <= 1) {
    return false;
  }
  for (ll i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}

int main() {
  int a, b;
  cin >> a >> b;
  vector<ll> aa = enumerateDivisors(a);
  vector<ll> bb = enumerateDivisors(b);
  vector<ll> common;
  for (ll aaa : aa) {
    if (find(bb.begin(), bb.end(), aaa) != bb.end()) {
      common.push_back(aaa);
    }
  }

  ll res = 1; // 1の分
  for (ll x : common) {
    if (isPrime(x) || x == a || x == b) {
      res++;
    }
  }
  cout << res << endl;

  return 0;
}