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>
typedef int value_type;

value_type med3(value_type x, value_type y, value_type z)
{
	if (x < y)
		if (y < z) return y; else if (z < x) return x; else return z; else
			if (z < y) return y; else if (x < z) return x; else return z;
}

void quicksort(value_type a[], int left, int right)
{
	if (left < right) {
		int i = left, j = right;
		value_type tmp, pivot = med3(a[i], a[i + (j - i) / 2], a[j]); /* (i+j)/2ではオーバーフローしてしまう */
		while (1) { /* a[] を pivot 以上と以下の集まりに分割する */
			while (a[i] < pivot) i++; /* a[i] >= pivot となる位置を検索 */
			while (pivot < a[j]) j--; /* a[j] <= pivot となる位置を検索 */
			if (i >= j) break;
			tmp = a[i]; a[i] = a[j]; a[j] = tmp; /* a[i],a[j] を交換 */
			i++; j--;
		}
		quicksort(a, left, i - 1);  /* 分割した左を再帰的にソート */
		quicksort(a, j + 1, right); /* 分割した右を再帰的にソート */
	}
}

int main()
{
	
	int N,i;
	long int c[10000], ans=0;
	scanf("%d", N);
	for (i = 0; i < N; i++)
	{
		scanf("%ld", c[i]);
	}
	quicksort(c, 0, N);
	if (c[N / 2] != c[(N / 2) + 1])
	{
		ans=c[(N / 2) + 1] - c[N / 2];
	}
	printf("%ld", ans);
}