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>
using namespace std;

bool tab[100000005];
vector<int> pierwsze;

void make(long long a)
{
    for (int i = 2; i <= a; i++)
        if (!tab[i])
        {
            pierwsze.push_back(i);
            for (int j = 2; j * i <= a; j++)

                tab[j * i] = true;
        }
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    long long a, b;
    cin >> a >> b;
    make(sqrt(max(a,b)));
    int ans = 1;
    for (auto u : pierwsze)
    {
        if (a % u == 0 && b % u == 0)
            ans++;
        if (a < u && b < u)
            break;
    }
    cout <<(ans>2||(a==1||b==1)?ans:(a==b?2:1));
    return 0;
}