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;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        var input = Console.ReadLine().Split(' ');

        var gcp = EuclideanAlgorithm(long.Parse(input[0]), long.Parse(input[1]));

        var primeNum = CountPrimeFactor(gcp);

        Console.WriteLine(primeNum + 1);

        Console.ReadLine();


    }

    public static long EuclideanAlgorithm(long a, long b)
    {
        if (a < b)
        {
            var temp = a;
            a = b;
            b = temp;
        }

        long result = 0;
        while (true)
        {
            var remainder = a % b;
            if (remainder == 0)
            {
                result = b;
                break;
            }
            else
            {
                a = b;
                b = remainder;
            }
        }

        return result;
    }

    public static int CountPrimeFactor(long max)
    {
        var primeList = new HashSet<long>();

        var n = max;

        for (var i = 2; i * i <= max; i++)
        {
            while (n % i == 0)
            {
                primeList.Add(i);
                n /= i;
            }
        }
        if(n > 1)
        {
            primeList.Add(n);
        }
        return primeList.Count;
    }
}