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 = 0;

        if (0 < a) count = 1;
        if ((a % 2 == 0) && (b % 2 == 0)) count = 2;

        for (long i = 3; i <= a; i += 2)
        {
            if((a % i == 0) && (b % i == 0))
            {
                if (IsPrime(i) == true) count++;
            }
        }

        Console.WriteLine(count);
    }

    public static bool IsPrime(long num)
    {
        if (num < 2) return false;
        else if (num == 2) return true;
        else if (num % 2 == 0) return false;

        double sqrtNum = Math.Sqrt(num);
        for (long i = 3; i <= sqrtNum; i += 2)
        {
            if (num % i == 0)
            {
                return false;
            }
        }
        return true;
    }
}