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.Scanner;


public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] d = new int[n+1];
		for(int i=0;i<n;i++) {
			d[i+1] = sc.nextInt();
		}
		d[0] = 0;
		for(int i=12;i>=0;i--) {
			if (check(n+1,d,i)) {
				System.out.println(i);
				return;
			}
		}
		System.out.println(0);
	}

	public static boolean check(int n,int[] d, int k) {
		int[][] m = new int[n][n];
		for(int i=0;i<n;i++) {
			for(int j=0;j<n;j++) {
				if (i == j) continue;
				boolean same = Math.abs(d[i]-d[j]) >= k;
				boolean diff = Math.min(d[i] + d[j], 24 - d[i] - d[j]) >= k;
				if (!same && !diff) {
					return false;
				}
				if (same && diff) {

				}else if(same){
					m[i][j] = m[j][i] = 1;
				}else{
					m[i][j] = m[j][i] = -1;
				}
			}
		}
//		System.out.println(Arrays.deepToString(m));
		int[] sgn = new int[n];
		boolean ok = true;
		try {
			for(int i=0;i<n;i++) {
				if (sgn[i] == 0) {
					sgn[i] = 1;
					dfs(i,n,m,sgn);
				}
			}
		}catch (RuntimeException e) {
			ok = false;
		}
		return ok;

		//return (a <= i && i < b) || (a <= i + 24 && i + 24 < b);
	}

	public static void dfs(int v, int n, int[][] m, int[] sgn) {
		for(int u=0;u<n;u++) {
			if (v==u || m[v][u] == 0) continue;
			int sgn2 = sgn[v] * m[v][u];
			if (sgn[u] != 0) {
				if (sgn[u] != sgn2) {
					//fail
					throw new RuntimeException();
				}
			}else{
				sgn[u] = sgn2;
				dfs(u,n,m,sgn);
			}
		}
	}


}