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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
#define repr(i, n) for (int i = n; i >= 0; i--)
#define INF 1e9
using namespace std;

long int A, B;

int gcd(int a, int b)
{
    while (b)
    {
        int m = a % b;
        a = b;
        b = m;
    }
    return a;
}

bool IsPrime(int 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()
{

    cin >> A >> B;

    long int target = gcd(A, B);

    vector<long int> C;

    long int head = 1, tail = target;
    while (head < tail)
    {
        if (target % head == 0)
        {
            C.push_back(head);

            if (head != target / head)
            {
                C.push_back(target / head);
            }
            tail = target / head;
        }
        head++;
    }

    // for (int i = 1; i <= target; i++)
    // {
    //     if (A % i == 0 && B % i == 0)
    //     {
    //         C.push_back(i);
    //     }
    // }

    // rep(i, C.size())
    // {
    //     cout << C[i] << " ";
    // }

    long int count = 1;
    rep(i, C.size())
    {
        if (IsPrime(C[i]))
        {
            count++;
        }
    }
    cout << count << endl;
}