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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
//ユーグリッドの互除法による最大公約数を求めるメソッド
ll gcd(ll a, ll b) {
  ll smaller = min(a, b);
  ll bigger = max(a, b);
  ll surplus = bigger % smaller;
  if (surplus == 0) return smaller;
  surplus = gcd(smaller, surplus);
  return surplus;
}

//最小公倍数を求めるメソッド
ll lcm(ll a, ll b) {
  return a * b / gcd(a, b);
}

//素数判定
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
    double sqrtNum = sqrt(num);
    for (int 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() {
  ll a, b;
  cin >> a >> b;
  ll n = gcd(a, b);
  ll ans = 1;
  if(n % 2 == 0) ans++;
  for (ll i = 3; i <= n; i+= 2){
    if(n % i != 0) continue;
    if(isPrime(i)) ans++;
  }
  cout << ans << endl;
  return 0;
}