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.

Examples

Input

3
7 12 8


Output

4


Input

2
11 11


Output

2


Input

1
0


Output

0
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Main implements Runnable {
	public static void main(String[] args) {
		new Main().run();
	}

	long MOD = (long) (1e9 + 7);

	public void run() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] D = new int[N];
		for (int i = 0; i < N; ++i) {
			D[i] = sc.nextInt();
		}
		boolean[][] ok = new boolean[2][1 << 24];
		ok[0][1] = true;
		for (int i = 0; i < N; ++i) {
			int idx = i % 2;
			Arrays.fill(ok[idx ^ 1], false);
			for (int j = 0; j < 1 << 24; ++j) {
				if (!ok[idx][j])
					continue;
				if ((j & (1 << D[i])) == 0)
					ok[idx ^ 1][j | 1 << D[i]] = true;
				if ((j & (1 << ((24 - D[i]) % 24))) == 0)
					ok[idx ^ 1][j | 1 << ((24 - D[i]) % 24)] = true;
			}
		}
		int cur = N % 2;
		int ans = 0;
		for (int i = 0; i < 1 << 24; ++i) {
			if (ok[cur][i]) {
				int tmp = Integer.MAX_VALUE / 3;
				for (int j = 0; j < 24; ++j) {
					if ((i >> j) % 2 == 0)
						continue;
					int k = j + 1;
					while (k < 24 && (i >> k) % 2 == 0)
						++k;
					if (k == 24) {
						k = 0;
						while ((i >> k) % 2 == 0)
							++k;
						tmp = Math.min(tmp, Math.min(j - k, 24 - (j - k)));
					} else {
						tmp = Math.min(tmp, Math.min(24 - (k - j), k - j));
						j = k - 1;
					}
				}
				ans = Math.max(ans, tmp);
			}
		}
		System.out.println(ans);
	}

	static void tr(Object... objects) {
		System.out.println(Arrays.deepToString(objects));
	}
}
