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

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int [] d = new int[n + 1];
        d[0] = 0;
        for (int i = 1; i <= n; ++i) {
            d[i] = scanner.nextInt();
        }
        Arrays.sort(d);
        for (int i = 0; i <= n; ++i) {
            if (i % 2 == 0) {
                d[i] = 24 - d[i];
            }
        }
        int min = 24;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j <= n; ++j) {
                int diff = Math.abs(d[i] - d[j]);
                int current = Math.min(24 - diff, diff);
                min = Math.min(min, current);
            }
        }

        System.out.println(min);
    }
}
