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 AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            var n = int.Parse(Console.ReadLine());
            var vals = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).OrderBy(x => x).ToArray();
            // 中央値
            var median = vals[n / 2];
            var tmp = median;
            var result = 0;
            while(true)
            {
                var arc = vals.Where(x => x >= tmp).Count();
                var abc = vals.Where(x => x < tmp).Count();
                if(arc == abc)
                {
                    result++;
                }
                else if(arc > abc)
                {
                    break;
                }
                tmp--;
            }
            tmp = median + 1;
            while (true)
            {
                var arc = vals.Where(x => x >= tmp).Count();
                var abc = vals.Where(x => x < tmp).Count();
                if (arc == abc)
                {
                    result++;
                }
                else if (abc > arc)
                {
                    break;
                }
                tmp++;
            }
            Console.WriteLine(result);
        }
    }
}