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 swap(int x[],int i,int j){
  int tmp;
  tmp=x[i];
  x[i]=x[j];
  x[j]=tmp;
}

void Qsort(int x[],int left,int right){
  int pivot;
  int i,j;
  i=left;
  j=right;
  pivot=x[(left+right)/2];
  
  while(1){
    while(x[i]<pivot) i++;
    while(x[j]>pivot) j--;
    if(i>=j) break;
    
    swap(x,i,j);
    i++;
    j--;
  }
  
  if(left<i-1)
    Qsort(x,left,i-1);
  if(right>j+1)
    Qsort(x,j+1,right);
  
}
  
int main(){
  int n,tmp;
  
  
  scanf("%d",&n);
  int d[n];
  for(int i=0;i<n;i++){
    scanf("%d",&d[i]);
  }
  
  Qsort(d,0,n-1);
  
  int b;
  b=n/2;
  printf("%d",d[b]-d[b-1]);
  
  return 0;
}