Problem:
Score : 500 points

Problem Statement

In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi.

Takahashi checked and found that the time gap (defined below) between the local times in his city and the i-th person's city was D_i hours.
The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours.
Here, we are using 24-hour notation.
That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example.

Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours.

Find the maximum possible value of s.

Constraints

1 \leq N \leq 50

0 \leq D_i \leq 12

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N

Output

Print the maximum possible value of s.

Sample Input 1

3
7 12 8

Sample Output 1

4

For example, consider the situation where it is 7, 12 and 16 o'clock in each person's city at the moment when it is 0 o'clock in Takahashi's city. In this case, the time gap between the second and third persons' cities is 4 hours.

Sample Input 2

2
11 11

Sample Output 2

2

Sample Input 3

1
0

Sample Output 3

0

Note that Takahashi himself is also a participant.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        try (Scanner sc = new Scanner(System.in)) {
            int N = sc.nextInt();
            int[] D = new int[N + 1];
            D[0] = 0;
            for (int i = 0; i < N; i++) {
                D[i + 1] = sc.nextInt();
            }

            Arrays.sort(D);

            int minmax = (int) 1e9;
            int[] isClockWise = new int[N + 1];
            for (int i = 1; i < N + 1; i++) {
                int max = 0;
                int maxK = -1;
                for (int k = 0; k < 2; k++) {
                    int di = k == 0 ? D[i] : (24 - D[i]);
                    int min = (int) 1e9;
                    for (int j = 0; j < i; j++) {
                        int dj = isClockWise[j] == 0 ? D[j] : (24 - D[j]);
                        int diff = Math.abs(di - dj);
                        diff = Math.min(diff, 24 - diff);
                        min = Math.min(min, diff);
                    }
                    if (min > max) {
                        max = min;
                        maxK = k;
                    }
                }
                isClockWise[i] = maxK;
                minmax = Math.min(minmax, max);
            }
            System.out.println(minmax);
        }
    }
}