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: Accepted)

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

namespace ConsoleApp2
{


    public class Program
    {
        public static void Main(string[] args)
        {
            string[] ABString = Console.ReadLine().Split(' ');
            long A = long.Parse(ABString[0]);
            long B = long.Parse(ABString[1]);
            long ans = 1;

            long gcd = Gcd(A, B);

            Dictionary<long, long> elements = FindElements(gcd);

            ans += elements.Count;
            
            Console.WriteLine(ans);
            Console.ReadLine();
        }


        static Dictionary<long,long> FindElements(long N)
        {
            long curN = N;
            Dictionary<long, long> elements = new Dictionary<long, long>();
           
            for (int i = 2; i <= (long)Math.Sqrt(N); i++)
            {
                while (curN%i==0)
                {
                    curN /= i;
                    if (elements.ContainsKey(i)) elements[i]++;
                    else elements.Add(i, 1);
                }
            }
            if(curN!=1)elements.Add(curN, 1);
            return elements;
        }


        /// <summary>
        /// 最大公約数
        /// </summary>
        /// <param name="a"></param>
        /// <param name="b"></param>
        /// <returns></returns>
        public static long Gcd(long a, long b)
        {
            if (a < b)
                // 引数を入替えて自分を呼び出す
                return Gcd(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }

    }


}