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: Time Limit Exceeded)

using System;

class DisjointSetofCommonnDivisors
{
    public static void Main(string[] args)
    {
        string[] str = Console.ReadLine().Split(' ');
        long a = long.Parse(str[0]);
        long b = long.Parse(str[1]);

        int count = 1;
        long gcd = Gcd(a,b);
        long tmp = gcd;

        for(int i = 2; i*i <= tmp; i++)
        {
            if (gcd % i == 0) count++;  
            while(gcd % i == 0)
            {
                gcd /= i;
            }
        }

        if (gcd != 1) count++;
        

        Console.WriteLine(count);
    }

    static long Gcd(long a, long b)
    {
        if (a > b) return Gcd(b, a);

        long r = b % a;

        while (r != 0)
        {
            b = a;
            a = r;
            r = b % a;
        }

        return a;
    }


}