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 <algorithm>
#include <cstdint>
#include <iostream>
#include <vector>

using namespace std;

using value_t = uint64_t;

struct Input
{
    value_t a;
    value_t b;
};

static Input read_input()
{
    Input input{};
    cin >> input.a >> input.b;
    return input;
}

static bool is_common_divisor(const Input& input, int n)
{
    return (input.a % n) == 0 && (input.b % n) == 0;
}

static vector<value_t> solve_brute_force(const Input& input)
{
    vector<value_t> divisors;
    divisors.push_back(1);

    const auto smaller = std::min(input.a, input.b);
    for (auto n = 2; n <= smaller; ++n) {
        if (!is_common_divisor(input, n)) continue;

        bool found = false;
        for (auto it = divisors.begin() + 1; it != end(divisors); ++it) {
            if ((n % *it) == 0) {
                found = true;
                break;
            }
        }
        if (found) continue;

        divisors.push_back(n);
    }

    return divisors;
}

static void solve(const Input& input)
{
    const auto divisors = solve_brute_force(input);
    cout << divisors.size() << endl;
}

static void setup()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
}

int main(int argc, char** argv)
{
    setup();

    const auto input = read_input();
    solve(input);

    return 0;
}