Problem:
Score : 300 points

Problem Statement

Takahashi made N problems for competitive programming.
The problems are numbered 1 to N, and the difficulty of Problem i is represented as an integer d_i (the higher, the harder).

He is dividing the problems into two categories by choosing an integer K, as follows:

A problem with difficulty K or higher will be for ARCs.

A problem with difficulty lower than K will be for ABCs.

How many choices of the integer K make the number of problems for ARCs and the number of problems for ABCs the same?

Problem Statement

2 \leq N \leq 10^5

N is an even number.

1 \leq d_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
d_1 d_2 ... d_N

Output

Print the number of choices of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 1

6
9 1 4 4 6 7

Sample Output 1

2

If we choose K=5 or 6, Problem 1, 5, and 6 will be for ARCs, Problem 2, 3, and 4 will be for ABCs, and the objective is achieved.
Thus, the answer is 2.

Sample Input 2

8
9 1 14 5 5 4 4 14

Sample Output 2

0

There may be no choice of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 3

14
99592 10342 29105 78532 83018 11639 92015 77204 30914 21912 34519 80835 100000 1

Sample Output 3

42685

Solution in C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

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

namespace ConsoleApp15
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.ReadLine(); //読み飛ばし

            string line = Console.ReadLine();

            int[] nums = line.Split(' ').Select(c => int.Parse(c)).ToArray();

            if (nums.Length % 2 == 1)
            {
                Console.WriteLine("0");
                return;
            }

            int count=0;
            for (int i = nums.Min(); i <= nums.Max(); i++)
            {
                if (Candevide(nums, i))
                {
                    count++;
                }
            }
            Console.WriteLine(count.ToString());
            
        }

        private static bool Candevide(int[] data , int threshold)
        {
            if (data.Length % 2 == 1)
            {
                return false;
            }

            return data.Where(num => num >= threshold).Count() == data.Count() / 2;
        }
    }
}