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>

#define swap(type,x,y) do{type t=x; x=y; y=t;}while(0)

void sort(int a[],int left,int right){
	int pl=left;
	int pr=right;
	int x=a[(pl+pr)/2];

	do{
		while(a[pl]<x) pl++;
		while(a[pr]>x) pr--;
		if(pl<=pr){
			swap(int,a[pl],a[pr]);
			pl++;
			pr--;
		}
	}while(pl<=pr);
	if(left<pr) sort(a,left,pr);
	if(pl<right) sort(a,pl,right);
}

int main(void){
	int i,j;

  int n;

  scanf("%d",&n);

  int d[n+1];

  for(i=1;i<=n;i++){
    scanf("%d",&d[i]);
  }

  sort(d,1,n);

  printf("%d\n",d[n/2+1]-d[n/2]);

	return 0;
}