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
#include <iostream>
using namespace std;
#define FAIL { cout << "0\n"; return 0; }
#define active(i,k) ((i)&(1<<(k)))

int main() {
    int N, t, a[24] = {}, i;
    cin >> N;
    a[0] = 1;
    for (i = 0; i < N; i++) {
        cin >> t;
        if (a[t]) {
            t = (24-t)%24;
            if (a[t]) FAIL;
        }
        a[t]++;
    }

    int R = 0;
    for (i = 0; i < 4096; i++) {
        int j = 0, k = 1, r = 24;

        for (int l = 1; l < 12; l++) {
            // if a[l] is inverted and a[24-l] exists
            if (active(i, l) && a[24-l]) r = 0;
        }

        // find next k that is ACTIVATED
        while (k < 24) {
            // k is activated if:
            // + k >= 12: a[k] exists OR inverted is active
            // + k < 12: a[k] exists AND not inverted

            for (;k < 24; k++) {
                if (k >= 12) {
                    if (a[k] || (a[24-k] && active(i, 24-k))) break;
                } else {
                    if (a[k] && !active(i, k)) break;
                }
            }
            if (k == 24) break;

            //printf("j,k: %d, %d\n", j, k);
            r = min(r, k-j);
            j = k++;
        }

        // check end
        r = min(r, 24-j);
        //printf("%d: %d\n", i, r);
        R = max(R, r);
    }
    cout << R << '\n';


}