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: Time Limit Exceeded)

#include <stdio.h>
int main(void){
	int y,arcomp=0;
	scanf("%d", &y);
	int arr[y];
	for(arcomp;arcomp<y;++arcomp){
		scanf("%d", &arr[arcomp]);
	}                       
	int min,troca,i;
	for(arcomp=0;arcomp<y-1;++arcomp){
		min=arcomp;
		for(i=arcomp+1;i<y;++i){
			if (arr[i]<arr[min]){
				min = i;
			}
		}
		if(arr[arcomp] != arr[min]){
			troca = arr[arcomp];
			arr[arcomp]= arr[min];
			arr[min] = troca;
		}
	}
	printf("%d\n",arr[y/2]-arr[y/2-1]);
	return 0;

}