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;

class Program {
    static void Main(string[] args) {
        //入力
        int N = int.Parse(Console.ReadLine());

        string s = Console.ReadLine();
        //int[] d = new int[N];
        //変数の割り当て
        //for (int i = 0; i < N; i++) {
        //    d[i] = int.Parse(s.Split(' ')[i]);
        //}
        //配列のソート
        //Array.Sort(d);

        List<int> list = new List<int> { };
        for (int i = 0; i < N; i++) {
            list.Add(int.Parse(s.Split(' ')[i]));
        }
        //リストのソート
        list.Sort();

        //B - Aの値
        int ans = 0;
        //検証
        //ans = d[N / 2] - d[(N / 2) - 1];
        ans = list[N / 2] - list[(N / 2) - 1];

        //出力
        Console.WriteLine(ans);
    }
}