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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

// クイックソート用の昇順関数
int cmpnum(const void * n1, const void * n2) 
{
	if (*(long *)n1 > *(long *)n2)
		return 1;
	else if (*(long *)n1 < *(long *)n2)
		return -1;
	else
		return 0;
}

int main(void)
{
	long N;
	scanf("%ld", &N);
	long d[N];
	long count = 0;
	long arc = N / 2;
	long abc = (N / 2) - 1;
	long i = 0;
	while (i < N)
	{
		scanf("%ld", &d[i]);
		i++;
	}
	qsort(d, sizeof(d) / sizeof(d[0]), sizeof(long), cmpnum);
	printf("%ld\n",d[arc] - d[abc]);
}