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>
#include <math.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define lper(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9;
const ll INF=4e18;
const ll dy[4]={1,0,-1,0};
const ll dx[4]={0,-1,0,1};
int main() {
 ll N;
  cin >>N;
  vl vec(N);
  rep(i,N) {
    cin >>vec.at(i);
  }
  
  sort(all(vec));
  
  double mid=(double)(vec.at((N-1)/2)+vec.at(N/2))/(double)2;
  ll a=vec.at((N-1)/2);
  ll b=vec.at(N/2);
  if(a==b) {
    cout<<0<<endl;
  }
  
  else {
    ll count=0;
    while(b>a) {
      b=b-1;
      count++;
    }
    cout<<count<<endl;
    
    
  }
  
  
}