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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ALL(v) v.begin(), v.end()
#define RALL(a) (a).rbegin(), (a).rend()
#define SORT(c) sort((c).begin(),(c).end())
#define dbug(x) cout<<#x<<"="<<x<<endl
#define m0(x) memset(x,0,sizeof(x))
#define rep(i, n) for(int i = 0; i < n; i++)
#define repr(i, n) for(int i = n; i >= 0; i--)
#define reps(i, m, n) for(int i = m; i < n; i++)
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))

using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pii;
typedef vector<pii> VP;
typedef long long ll;
typedef pair<ll, ll> pl;

template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
  for(T& x: vec) is >> x;
  return is;
}

int main(){

  int N;
  cin >> N;

  vector<int> d(N);

  cin>>d;

  SORT(d);

  int mn,mx;
  int sum;

  mn=d[(N/2)-1];
  mx=d[(N/2)];

  cout << mx-mn <<endl;
  return 0;
}