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++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS

/* include ***********************/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

/* define *************************/
//for
#define REP(i,n) for(int i=0,i##_len=(int)(n);i<i##_len;i++)
#define REPS(i,n) for(int i=1,i##_len=(int)(n);i<=i##_len;i++)
#define RREP(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define RREPS(i,n) for(int i=(int)(n);i>0;i--)
#define FOR(i,s,n) for(int i=(int)(s),i##_len=(int)(n);i<i##_len;i++)
#define RFOR(i,s,n) for(int i=(int)(n)-1,i##_len=(int)(s);i>=i##_len;i--)

//printf
#define PRINTD(d) printf("%d\n",(d))
#define PRINTL(d) printf("%lld\n",(d))

//memset
#define m0(s) memset(s,0,sizeof(s))
#define ml(s) memset(s,63,sizeof(s))
#define fill(s,c) memset(s,c,sizeof(s))

#define INF 1e9
#define MOD 1000000007

typedef long long ll;
typedef unsigned long long ull;

int diff[4][2] = {
	{ 0, -1 },
	{ -1, 0 },
	{ 1, 0 },
	{ 0, 1 },
};
//今回の変数
int d[100000];

int Min(int a, int b) {
	return (a)<(b) ? (a) : (b);
}
int Max(int  a, int  b) {
	return (a) > (b) ? (a) : (b);
}
ll Minl(ll a, ll b) {
	return (a)<(b) ? (a) : (b);
}
ll Maxl(ll a, ll b) {
	return (a)>(b) ? (a) : (b);
}
void ShowData(int x[], int left, int right)
{
	int i;
	for (i = left; i <= right; i++)printf("%d  ", x[i]);
	printf("\n");
}
void hSwap(int x[], int i, int j)
{
	int temp;

	temp = x[i];
	x[i] = x[j];
	x[j] = temp;
}
void QSort(int x[], int left, int right, int n) {
	int i, j;	//左端,右端
	int pivot;	//軸

	i = left;
	j = right;
	pivot = x[(left + right) / 2];

	while (1) {
		if (n > 0) {				//n>0なら昇順、n<=0なら降順
			while ((x[i] < pivot) && (i <= right))i++;	//軸値より大きい要素
			while ((pivot < x[j]) && (i <= right))j--;	//軸値より小さい要素
		}
		else {
			while ((x[i] > pivot) && (i <= right))i++;	//軸値より小さい要素
			while ((pivot > x[j]) && (i <= right))j--;	//軸値より大きい要素
		}
		if (i >= j)	break;
		hSwap(x, i, j);
		i++; j--;
	}
	//ShowData(x, left, right);

	if (left < i - 1)QSort(x, left, i - 1, n);
	if (j + 1 <  right)QSort(x, j + 1, right, n);
}

int main() {
	int n;
	scanf("%d", &n);

	REP(i, n) {
		scanf("%d", &d[i]);
	}
	QSort(d, 0, n - 1, 1);

	PRINTD(d[n / 2] - d[n / 2 - 1]);
}