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 (Clang 3.8.0) (judge verdict: Accepted)

//#include <boost/multiprecision/cpp_dec_float.hpp>
//#include <boost/multiprecision/cpp_int.hpp>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <math.h>
#include <algorithm>
using namespace std;
//namespace mp = boost::multiprecision;
//using llll = mp::cpp_int;
// 仮数部が1024ビットの浮動小数点数型(TLEしたら小さくする)
//using Real = mp::number<mp::cpp_dec_float<1024>>;
using ll = long long;
const double EPS = 1e-10;
const ll MOD = 7+(1e+9);

int main(){
  ll n;
  cin>>n;
  vector<ll>v(n,0);
  for(ll i=0;i<n;i++){
    cin>>v[i];
  }
  sort(v.begin(),v.end());
  ll ans=v[n/2]-v[n/2-1];
  cout<<ans<<endl;
}