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 SZ(x) ((int)(x).size())
typedef long long ll;
typedef long double ld;
using namespace std;

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); cout.setf(ios::fixed); cout.precision(20);
    ll A, B; cin >> A >> B;
    ll g = __gcd(A, B);
    vector<ll> a;
    for (ll i = 1; i * i <= g; i++) {
        if (g % i == 0) {
            for (ll j : {i, g / i}) {
                if (a.empty() || j != a.back()) {
                    a.push_back(j);
                }
            }
        }
    }
    sort(a.begin(), a.end());
    map<ll, bool> used;
    ll res = 0;
    for (ll x : a) {
        bool good = true;
        for (ll i = 1; i * i <= x && good; i++) {
            if (x % i == 0) {
                for (ll j : {i, g / i}) {
                    if (used[j] && j > 1) {
                        good = false;
                    }
                }
            }
        }
        if (good) {
            res++;
            used[x] = true;
        }
    }
    cout << res << "\n";
}