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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
typedef long long ll;

ll n;
ll s[100005];
ll cnt;

int main()
{
    scanf("%lld", &n);
    for (int i = 0; i < n; i++) {
        ll d;

        scanf("%lld", &d);

        s[d]++;
    }

    for (int i = 1; i < 100005; i++) {
        s[i] = s[i - 1] + s[i];
    }

    for (int k = 1; k <= 100000; k++) {
        if (s[k] - s[0] == s[100000] - s[k]) {
            cnt++;
        }
    }

    printf("%lld\n", cnt);
    
    return 0;
}