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

#define REP(i,n) for(int i = 0; i < n; i++)
#define ll long long
#define PI M_PI
#define bf printf("\n");

void swap(int *x, int *y){
    int tmp;
    tmp = *x;
    *x = *y;
    *y = tmp;
}

void Bsort(int array[], int N){
    int i, j;
    for(j = 1; j < N; j++){
        for(i = 0; i < N-j; i++){
            if(array[i] > array[i+1]) swap(&array[i], &array[i+1]);
        }
    }
}

int main(void){
    int N;  scanf("%d",&N);
    int d[N];
    REP(i,N) scanf("%d",&d[i]);

    Bsort(d,N);
    printf("%d",d[N/2]-d[N/2-1]);
    return 0;
}