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;

//using Matrix = vector< vector<int> >;
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1 << 30;
int dx[5] = {0, 0, 1, -1, 0}, dy[5] = {1, -1, 0, 0, 0};
bool cmp(P a, P b) { return a.second < b.second; } //End sort (Interval scheduling problem)

int n;
vector<int> d;
int ans;

int main(void){
	cin>>n;
	d.resize(n);
	for (int i = 0; i < n; i++) cin>>d[i];

	sort(d.begin(), d.end());
	for (int i = 0; i < 100010; i++) {
		auto ite = upper_bound(d.begin(), d.end(), i);
		int abc = 0, arc = 0;
		abc = distance(d.begin(), ite);
		arc = distance(ite, d.end());

		if (abc == arc) ans++;
	}

	cout<< ans <<endl;
	
	return 0;
}