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;

namespace ABC142D
{
    class Program
    {
        static void Main(string[] args)
        {
            var inputArr = Console.ReadLine().Split().Select(long.Parse).ToArray();
            //var inputInt = int.Parse(Console.ReadLine());

            long A = inputArr[0];
            long B = inputArr[1];

            //公約数の算出
            List<long> kouyakusuu = new List<long>();
            kouyakusuu.Add(1);
            double root = Math.Min(Math.Sqrt(A), Math.Sqrt(B));
            kouyakusuu.AddRange(Kouyakusuu(A, B, 2, root));

            //int ans = 0;
            //string ans = "";

            Console.WriteLine(kouyakusuu.Distinct().Count());
            Console.ReadKey();
        }

        static List<long> Kouyakusuu(long A, long B, long atai, double root)
        {

            for (long i = atai; i < root; i++)
            {
                //割り切れた場合
                if (A % i == 0 && B % i == 0)
                {
                    var ans = Kouyakusuu(A / i, B / i, atai, root);
                    ans.Add(i);
                    return ans;
                }
                else
                {

                }
            }

            return new List<long>();
        }
    }
}