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

namespace AtCoder.D
{
    public class Program
    {
        public static void Main() { var r = GetResult(); Debug.WriteLine(r); Console.Write(r); }

        private static object GetResult()
        {
            var inputs = ReadLongs();
            var A = inputs[0];
            var B = inputs[1];

            var min = Math.Min(A, B);
            var max = Math.Max(A, B);
            var nums = new List<long>();
            for (long i = 1; i <= min; i++)
            {
                nums.Add(i);
            }

            var count = 0L;
            while (true)
            {
                var i = nums[0];
                if (min % i == 0 && max % i == 0)
                {
                    count++;
                }

                nums.RemoveAt(0);
                if (nums.Count == 0) break;

                if (i == 1) continue;

                // おしりから消していく
                for (long j = nums.Count - 1; j != 0; j--)
                {
                    if (nums[(int)j] % i == 0)
                    {
                        nums.RemoveAt((int)j);
                    }
                }
            }

            return count;
        }

        #region Console

        public static string ReadText() { return Console.ReadLine(); }
        public static List<string> ReadTexts() { return Console.ReadLine().Split(' ').ToList(); }
        public static long ReadLong() { return long.Parse(Console.ReadLine()); }
        public static List<long> ReadLongs() { return Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToList(); }

        #endregion
    }
    public class EratosthenesSieve
    {
        public List<long> Primes { get; private set; }
        public Dictionary<long, bool> IsPrime { get; private set; }

        public EratosthenesSieve(long max)
        {
            Primes = new List<long>();
            IsPrime = new Dictionary<long, bool>();
            for (long i = 1; i <= max; i++)
            {
                IsPrime.Add(i, false);
            }

            for (long i = 2; i <= max; i++)
            {
                IsPrime[i] = true;
            }

            for (long i = 2; i <= max; i++)
            {
                if (IsPrime[i])
                {
                    Primes.Add(i);
                    for (long j = 2 * i; j <= max; j += i)
                    {
                        IsPrime[j] = false;
                    }
                }
            }
        }
    }
}