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: Runtime Error)

#include<iostream>
#include<algorithm>
using namespace std;
int vis[1000005];
long long gcd(long long x, long long y)
{
    return y == 0 ? x : gcd(y, x % y);
}
void solve()
{
    vis[1] = 0;
    vis[2] = 0;
    for(long long i = 2;i <= 1000005;i++)
    {
        if(vis[i] == 0)
        {
            long long t = 2;
            for(long long j = t * i;j <= 1000005;j = t * i)
            {
                vis[j] = 1;
                t++;
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    long long a,b;
    solve();
    cin >> a >> b;
    long long T = gcd(a, b);
    long long cnt = 1;
    for(long long i = 2;i <= T;i++)
    {
        if(T % i == 0&& vis[i] == 0)
        {
            //cout << i << endl;
            cnt++;
        }
    }
    cout << cnt << endl;
    return 0;
}