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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp5
{
    class Program
    {
        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 (int i = 3; i <= sqrtNum; i += 2)
            {
                if (num % i == 0)
                {
                    return false;
                }
            }
            return true;
        }

        static void Main(string[] args)
        {
            String S = Console.ReadLine();
            String[] L = S.Split(' ');
            long A = long.Parse(L[0]);
            long B = long.Parse(L[1]);

            if (A < B)
            {
                long t = A;
                A = B;
                B = t;
            }

            long R = A % B;
            while (R != 0)
            {
                A = B;
                B = R;
                R = A % B;
            }

            int ans = 1;

            for (int i = 1; i * i * 2<= B; i++)
            {
                if (B % i == 0)
                {
                    if(IsPrime(i) == true)
                    {
                        ans++;
                        if (i != B / i && IsPrime(B / i) == true)
                        {
                            ans++;
                        }
                    }

                }
            }


            Console.WriteLine(ans);



        }
    }
}