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;

typedef  long long ll;
template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b;return true;}return false;}

#define rep(i,n) for(int i=0;i<(int)n;i++)
#define rep2(i, begin_i, end_i) for (int i = (int)begin_i; i < (int)end_i; i++)

long long INF = 1LL<<60;
long long dp[100010];
int h[100010];
int n;
int main(){
    cin>>n;
    vector<int> d(n);
    rep(i,n)cin>>d.at(i);
    if(n%2){
        cout<<"0"<<endl;
    }
    sort(d.begin(),d.end());

    int right_smallest=n/2,left_upper=right_smallest-1;
    cout<<d.at(right_smallest)-d.at(left_upper)<<endl;
    return 0;
}