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)

// AtCoder ABC132: C - Divide the Problems
// 2019.9.3 bal4u

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

#if 1
#define gc() getchar_unlocked()
#else
#define gc() getchar()
#endif

int in() {   // 非負整数の入力
	int n = 0, c = gc();
	do n = 10 * n + (c & 0xf); while ((c = gc()) >= '0');
	return n;
}

#define MAX 100000
int d[MAX+3];

int cmp(const void *u, const void *v) { return *(int *)u - *(int *)v; }

int main()
{
	int i, k, N, ans;

	N = in(); for (i = 0; i < N; i++) d[in()]++;
	N >>= 1;
	k = 0; for (i = 0; k < N; i++) k += d[i];
	if (k > N) ans = 0;
	else {
		k = i-1; while (d[i] == 0) i++;
		ans = i-k;
	}
	printf("%d\n", ans);
	return 0;
}