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 <iostream>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
template <typename T = int>
T in() {
    T temp; cin >> temp; return temp;
}

template <typename VType = int>
list<VType> const LR(VType start, VType end) {
    list<VType> ret; for (VType i = start; i < end; i++) ret.push_back(i); return move(ret);
}

int main() {
    int const n(in());
    multiset<int> d;

    for (int i = 0; i < n; i++) {
        int temp(in());
        d.insert(temp);
    }

    multiset<int>::iterator ite = d.cbegin();
    for (int i = 0; i < (n/2 -1 ); i++) {
        
        ite++;
    }
    auto next = ite;
    next++;
    cout << *next - *ite << endl;


    return 0;
}