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>


void insertionSort(int a[],int n){
    int i,j,v;
    for(i=1;i<n;i++){
        v=a[i];          //v:未ソート部分の先頭
        j=i-1;
        while(j>=0&&a[j]>v){
            a[j+1]=a[j];    //vより大きい要素を後方へ移動
            j--;
        }
        a[j+1]=v;
    }
}

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