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;

#define rep(i , j , n) for(long long i = j ; i <= n ; i++)
#define per(i , j , n) for(long long i = j ; i >= n ; i--)
const ll N = 123456;

ll a[N];
int big[N];
int small[N];
int ans = 0;
const ll mod = 1e9 + 7;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n;
    cin >> n;

    rep(i , 1 , n) cin >> a[i];

    ll maxi = INT_MIN;

    rep(i , 1 , n) maxi = max(maxi , a[i]);

    rep(i , 1 , n) {
        big[a[i]]++;
        small[a[i]]++;
    }

    per(i , maxi , 1) {
        big[i] += big[i + 1];
    }

    rep(i , 1 , maxi) {
        small[i] += small[i - 1];
    }

    rep(i , 1 , maxi) {
        if(small[i - 1] == big[i]) {
           // cout << i << " " << big[i] << " " << small[i - 1] << endl;
            ans++;
        }
    }

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