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>
#include <stdlib.h>
#define N 100000

void swap (int *x, int *y) {
  int temp;    // 値を一時保存する変数

  temp = *x;
  *x = *y;
  *y = temp;
}

/* 選択ソート */
void selection_sort (int array[], int array_size) {
  int i, j, min_index;

  for (i = 0; i < array_size-1; i++) {
    min_index = i;   // 先頭要素が一番小さいとする
    for (j = i + 1; j < array_size; j++) {
      if (array[j] < array[min_index]) { min_index = j; }   // 値の比較、最小値の更新
    }
    swap(&array[min_index], &array[i]);   // 最小値と先頭要素を交換
  }
}

int main(void){
    int i,a[N],n;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    selection_sort(a,n);
    printf("\n%d\n",a[(n)/2]-a[(n-1)/2]);
    return 0;
}