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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;

class Program
{
    static void Main(string[] args)
    {
        var ab = Console.ReadLine().Split();
        var a = long.Parse(ab[0]);
        var b = long.Parse(ab[1]);

        var g = gcd(a, b);
        var ans = m1(g) + 1;

        Console.WriteLine(ans);
    }

    static long m1(long n)
    {
        var ans = 0L;
        for (long i = 2; i * i <= n; i++)
        {
            if (n % i != 0) continue;

            ans++;
            while (n % i == 0) n /= i;
        }

        if (n != 1) ans++;
        return ans;
    }

    static long gcd(long a, long b)
    {
        return b == 0 ? a : gcd(b, a % b);
    }
}