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: Runtime Error)

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

namespace Contest
{
    class ABC
    {
        static void Main()
        {
            Solve();
        }
        private static void Solve()
        {
            long a = long.Parse(Console.ReadLine());
            long b = long.Parse(Console.ReadLine());
            long gcd = Gcd(a, b);
            var res = Factorize(gcd);
            Console.WriteLine(res.Keys.Count + 1);
        }
        // ユークリッドの互除法 
        static long Gcd(long a, long b)
        {
            return b == 0 ? a : Gcd(b, a % b);
        }
        /// <summary>
        /// 素因数分解
        /// </summary>
        /// <param name="n"></param>
        /// <returns></returns>
        public static Dictionary<long, int> Factorize(long n)
        {
            var ret = new Dictionary<long, int>();
            for (long i = 2; i * i <= n; i++)
            {
                if (n % i != 0)
                {
                    continue;
                }

                ret[i] = 0;
                while (n % i == 0)
                {
                    n /= i;
                    ret[i]++;
                }
            }

            if (n > 1)
            {
                if (ret.ContainsKey(n))
                {
                    ret[n]++;
                }
                else
                {
                    ret[n] = 1;
                }
            }

            return ret;
        }
    }
}