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 sws ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ll long long
#define vi vector<int>
#define pb push_back
#define pii pair<int, int>
#define mp make_pair
#define ff first
#define ss second
#define all(x) x.begin(), x.end()
#define fo(i, n) for( int i = 0; i < n; i++ )


int main() {
    sws;

    int n;
    cin >> n;
    vi v(n);

    fo(i, n) {
        cin >> v[i];
    }

    sort(all(v));

    // fo(i, n) {
    //     cout << v[i] << " ";
    // }
    // cout << endl;

    cout << (v[n / 2] - v[n / 2 - 1]) << endl;

    return 0;
}