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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <cstdio>
#include <set>
#include <map>
#include <list>
#include <cstdlib>

using namespace std;

#define mod (1e9 + 7)
#define rep(i, n) for (int i = 0; i < n; i++)
#define all(x) x.begin(), x.end()

typedef long long ll;
ll gcd(ll a, ll b) {
    if(a%b) {
        return gcd(b, a%b);
    } else {
        return b;
    }
}
ll ans = 1;
bool is_prime(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;
}

set<ll> divisor(ll n, ll min)
{
    set<ll> res;
    for (int i = 1; i * i <= n; i++)
    {
        if (n % i == 0 && i*i <= min)
        {
            if(is_prime(i)) {
                ans ++;
            }

            if (n / i != i && (n / i) * (n / i) <= min)
            {
                if(is_prime(n/i)) {
                  ans ++;
                }
            }
        }
    }


    return res;
}

#include <math.h>



void solve1()
{
    ll a, b; cin >> a >> b;

    ll max = gcd(a, b);
    ll mina = min(a, b);
    set<ll> divs;
    divs = divisor(max, mina);

    cout << ans << endl;



}

int main()
{
    solve1();
}