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<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;

const int INF = 1001001001;
const int MOD = 1e9+7;

ll gcd(ll a, ll b) {
    if(b == 0) return a;
    return gcd(b, a%b);
}

vector<pair<ll, int>> factorize(ll n) {
    vector<pair<ll, int>> ret;
    for (ll i = 2; i*i <= n; i++)
    {
        if(n % i) continue;
        ret.emplace_back(i, 0);
        while (n%i == 0)
        {
            n /= i;
            ret.back().second++;
        }
    }
    if (n != 1) ret.emplace_back(n, 1);
    return ret;   
}

int main() {
    ll a, b; cin >> a >> b;
    ll g = gcd(a, b);
    auto f = factorize(g);
    int ans = f.size() + 1;
    cout << ans << endl;
    return 0;
}