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# (.NET Core 3.1.201) (judge verdict: Time Limit Exceeded)

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


class Program
{
    static void Main(string[] args)
    {
        // 入力
        //var s = Console.ReadLine();
        //var n = int.Parse(Console.ReadLine());
        long[] ab = Console.ReadLine().Split().Select(long.Parse).ToArray();
        var a = ab[0];
        var b = ab[1];
        long ans = 0;

        var max = Gcd(a, b);
        var set = new HashSet<long>();
        for (int i = 1; i <= max; i++)
        {
            if (a % i == 0 && b % i == 0) set.Add(i);
        }

        // bit全探索でABどちらで焼くかを決定する
        var n = set.Count();
        long[] array = set.ToArray();
        for (int i = 0; i < Math.Pow(2, n); i++)
        {
            var katei = new bool[n];
            for (int j = 0; j < n; j++)
            {
                var x = (i >> j) & 1;
                if (x == 1) katei[j] = true;
            }

            var ok = true;
            for (int j = 0; j < n; j++)
            {
                for (int k = j + 1; k < n; k++)
                {
                    if (!katei[j] && !katei[k])
                    {
                        var tmp = Gcd(array[j], array[k]);
                        if (tmp != 1)
                        {
                            ok = false;
                            break;
                        }
                    }
                }
                if (!ok) break;
            }

            if (ok)
            {
                var l = katei.Where(x => x == true).Count();
                ans = Math.Max(ans, n - l);
            }
        }
        // 出力
        Console.WriteLine(ans);
    }
    // ユークリッドの互除法 
    // 最大公約数を求める
    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;
    }
}