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: Accepted)

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)
        {
            var N = cin.ReadInt;
            var Number = new int[N];
            var ave = 0;

            for(var i=0;i<N;i++){
                Number[i] = cin.ReadInt;
                ave += Number[i];
            }
            Array.Sort(Number);

            var countUp = N/2 - 1;
            var countDown = N/2;
            var ans = 0;
            if(Number[countDown] != Number[countUp]) ans = Number[countDown] - Number[countUp];
            WriteLine(ans);
        }
    }

    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;}
    }

}