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 <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> P;


int main()
{
    cin.tie(0);
    ios::sync_with_stdio(0);

    int N;
    cin >> N;
    int count[13];
    for (int i = 0; i <= 12; i++) {
        count[i] = 0;
    }
    count[0]++;
    int d;
    for (int i = 0; i < N; i++) {
        cin >> d;
        count[d]++;
    }

    if (count[0] > 1) {
        cout << 0 << endl;
        return 0;
    }
    for (int i = 1; i <= 12; i++) {
        if (count[i] > 2) {
            cout << 0 << endl;
            return 0;
        }
    }

    int ans = 0;
    for (int i = 0; i < (1<<12); i++) {
        vector<int> time;
        time.push_back(0);
        for (int j = 1; j <= 12; j++) {
            if (count[j] == 1) {
                if (i & (1 << j))
                    time.push_back(j);
                else
                    time.push_back(24 - j);
            } else if (count[j] == 2) {
                time.push_back(j);
                time.push_back(24 - j);
            }
        }

        int mi = INT_MAX;
        for (int j = 0; j < time.size(); j++) {
            for (int k = j + 1; k < time.size(); k++) {
                int d = abs(time[j] - time[k]);
                mi = min(mi, min(d, 24 - d));
            }
        }

        ans = max(ans, mi);
    }

    cout << ans << endl;

    return 0;
}
