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>

void merge_sort (int array[], int left, int right) {
  int i, j, k, mid;
  int work[right+1];  // 作業用配列
  if (left < right) {
    mid = (left + right)/2; // 真ん中
    merge_sort(array, left, mid);  // 左を整列
    merge_sort(array, mid+1, right);  // 右を整列
    for (i = mid; i >= left; i--) { work[i] = array[i]; } // 左半分
    for (j = mid+1; j <= right; j++) {
      work[right-(j-(mid+1))] = array[j]; // 右半分を逆順
    }
    i = left; j = right;
    for (k = left; k <= right; k++) {
      if (work[i] < work[j]) { array[k] = work[i++]; }
      else                   { array[k] = work[j--]; }
    }
  }
}

int main(void){
  int n;
  scanf("%d", &n);
  int m[n];
  int i;
  for(i=0;i<n;i++){
    scanf("%d", &m[i]);
  }
  int count = 0;
  merge_sort(m,0,n-1);
  printf("%d", m[n/2]-m[n/2-1]);
}