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: Wrong Answer)

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

namespace ConsoleApp1
{
    class Program
    {
        static void Main(string[] args)
        {
            int num = int.Parse(Console.ReadLine());
            string[] val = Console.ReadLine().Split(' ');

            int[] intArray = val.Select(int.Parse).ToArray();
            Array.Sort(intArray);

            int count = 0;
            int max = intArray[intArray.Length - 1];

            List<int> before = new List<int>();
            List<int> after = new List<int>();

            for (int i = 0; i < max; i++)
            {
                before.Clear();
                after.Clear();

                for (int j = 0; j < num; j++)
                {
                    if (intArray[j] < i)
                        before.Add(intArray[j]);
                    else
                        after.Add(intArray[j]);
                }

                if (before.Count == after.Count)
                    count++;
            }

            Console.WriteLine(count.ToString());
        }
    }
}