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.Text;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using static System.Console;
using static System.Math;

namespace AtCoder
{
    public class Program
    {
        public static void Main(string[] args)
        {
            new Program().Solve(new ConsoleInput(Console.In, ' '));
        }

        public void Solve(ConsoleInput cin)
        {
            // 解答はココに書く
            var array = cin.ReadLongArray(2);
            var gcd = Gcd(array[0], array[1]);
            var max = gcd;
            long i = 2;
            var count = 1;
            var yakusuHash = new HashSet<long>();
            while (i * i <= max)
            {
                if (gcd % i == 0)
                {
                    if (yakusuHash.Add(i)) count++;
                    gcd = gcd / i;
                }
                else
                {
                    i++;
                }
            }
            WriteLine(count);
        }
        public long Gcd(long a, long b)
        {
            if (a > b) return Gcd(b, a);
            while (true)
            {
                var mod = b % a;
                if (mod == 0) return a;
                b = a;
                a = mod;
            }
        }
    }

    public class ConsoleInput
    {
        private readonly System.IO.TextReader _stream;
        private char _separator = ' ';
        private Queue<string> inputStream;
        public ConsoleInput(System.IO.TextReader stream, char separator = ' ')
        {
            this._separator = separator;
            this._stream = stream;
            inputStream = new Queue<string>();
        }
        public string Read
        {
            get
            {
                if (inputStream.Count != 0) return inputStream.Dequeue();
                string[] tmp = _stream.ReadLine().Split(_separator);
                for (int i = 0; i < tmp.Length; ++i)
                    inputStream.Enqueue(tmp[i]);
                return inputStream.Dequeue();
            }
        }
        public string ReadLine { get { return _stream.ReadLine(); } }
        public int ReadInt { get { return int.Parse(Read); } }
        public long ReadLong { get { return long.Parse(Read); } }
        public double ReadDouble { get { return double.Parse(Read); } }
        public string[] ReadStrArray(long N) { var ret = new string[N]; for (long i = 0; i < N; ++i) ret[i] = Read; return ret; }
        public int[] ReadIntArray(long N) { var ret = new int[N]; for (long i = 0; i < N; ++i) ret[i] = ReadInt; return ret; }
        public long[] ReadLongArray(long N) { var ret = new long[N]; for (long i = 0; i < N; ++i) ret[i] = ReadLong; return ret; }
    }
}