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: Runtime Error)

#include <stdio.h>

int main(void)
{
	int n, i, a, j, k;
	int Tmedi;
	int max[500];
	int min[500];
	int medi[1000];
	int d[1000];
	
	scanf("%d", &n);
	
	for (i = 0; i < n; i++){
		scanf("%d", &d[i]);
	}
	a = n / 2;
	max[0] = d[0];
	min[0] = d[0];

	for (i = 0; i < n; i++){
		if (max[0] < d[i]){
			max[0] = d[i];
		}
		else if (min[0] > d[i]){
			min[0] = d[i];
		}
		else if (max[0] > d[i] && min[0] < d[i]){
			medi[i] = d[i];
		}
	}
	j = 2;
	k = 1;
	while (1){
		for (i = 0; i < n - j; i++){
			if (medi[0] < medi[i] && medi[i] < max[i]){
				max[k] = medi[i];
			}
			else if (medi[0] > medi[i] && medi[i] > min[i]){
				min[k] = medi[i];
			}
		}
		j += 2;
		k++;
		if (k == a){
			break;
		}
	}
	printf("%d\n", max[k] - min[k]);
	
	return (0);
}