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>
using namespace std;

#define rep(i,n) for(int i = 0; i < ((int)(n)); i++)   //0-origin昇順
#define rep1(i,n) for(int i = 1; i <= ((int)(n)); i++) //1-origin昇順
#define rrep(i,n) for(int i = ((int)(n)-1); i >= 0; i--)  //0-origin降順
#define rrep1(i,n) for(int i = ((int)(n)); i >= 1; i--)   //1-origin降順
//rep2 -> 第二引数 m から n 回繰り返し ex) m=5 n=3 なら i=5,i=6,i=7 まで
#define rep2(i,m,n) for(int i = ((int)(m)); i < ((int)(n))+((int)(m)); i++)
#define SIZE(x) ((int)((x).size()))
#define all(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vii; //2次元配列
typedef vector<ll> vll;
typedef vector<string> vs;
template<typename T> bool chmax(T &a, const T &b){ if(a<b){ a=b; return 1;} return 0;}
template<typename T> bool chmin(T &a, const T &b){ if(a>b){ a=b; return 1;} return 0;}


/*FUNCs=================================================*/


/*MAIN==================================================*/
int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);   //cin cout 高速化

    int n; cin >> n;
    vi d(n); rep(i,n) cin >> d[i];
    sort(all(d));
    int itr1 = (SIZE(d))/2;
    int itr2 = itr1 - 1;
    int answer = d[itr1] - d[itr2];
    cout << answer << '\n';
}