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 <string>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <algorithm>
#include <deque>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <list>
#include <cstdio>
#include <iostream>
#include <cmath>
#include <climits>
#include <bitset>
#include <functional>
#include <numeric>
#include <ctime>
#include <cassert>
#include <cstring>
#include <fstream>
#include <iomanip>

using namespace std;

#define FOR(i, a, b) for(int (i)=(a);(i)<(b);(i)++)
#define IFOR(i, a, b) for(int (i)=(a);(i)<=(b);(i)++)
#define RFOR(i, a, b) for(int (i)=(a);(i)>=(b);(i)--)
#define REP(i, n) FOR((i), 0, (n))

using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

template<class T>
void print_container(T &container, string delim=" ") {
    bool first = true;
    for(auto &t: container){
        if(!first){
            cout << delim;
        }
        first = false;
        cout << t;
    }
    cout << endl;
}

int main() {
    int n;
    cin >> n;

    vector<int> d(n);
    REP(i, n)
        cin >> d[i];
    sort(d.begin(), d.end());
    int x1 = d[n/2-1], x2 = d[n/2];

    cout << x2 - x1 << endl;
    return 0;
}