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 <algorithm>
#include <cmath>
#include <iostream>
#include <limits>
#include <numeric>
#include <string>
#include <vector>

#define rep(i, x, n) for(int i = x; i < n; i++)
using namespace std;
const long long INF = 1LL << 60;

long long GCD(long long a, long long b) {
    if(b == 0)
        return a;
    else
        return GCD(b, a % b);
}
long long LCM(long long a, long long b) {
    long long g = GCD(a, b);
    return a / g * b;
}
int main() {
    int N;
    cin >> N;
    vector<int> d;
    rep(i, 0, N) {
        int a;
        cin >> a;
        d.push_back(a);
    }
    sort(d.begin(), d.end());
    if(d[N / 2] == d[N / 2 - 1]) {
        cout << 0 << endl;
    } else {
        cout << d[N / 2] - d[N / 2 - 1] << endl;
    }
}