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<math.h>
#include<stdlib.h>

void MergeSort(int x[ ], int left, int right)
{
    int mid, i, j, k;
    int temp[right+1];

    if (left >= right)              /* 配列の要素がひとつなら */
        return;                     /* 何もしないで戻る */

                                    /* ここでは分割しているだけ */
    mid = (left + right) / 2;       /* 中央の値より */
    MergeSort(x, left, mid);        /* 左を再帰呼び出し */
    MergeSort(x, mid + 1, right);   /* 右を再帰呼び出し */

      /* x[left] から x[mid] を作業領域にコピー */
    for (i = left; i <= mid; i++)
        temp[i] = x[i];

      /* x[mid + 1] から x[right] は逆順にコピー */
    for (i = mid + 1, j = right; i <= right; i++, j--)
        temp[i] = x[j];

    i = left;         /* i とj は作業領域のデーターを */
    j = right;        /* k は配列の要素を指している */

    for (k = left; k <= right; k++)    /* 小さい方から配列に戻す */
        if (temp[i] <= temp[j])        /* ここでソートされる */
            x[k] = temp[i++];
        else
            x[k] = temp[j--];
}

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

}