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;

using ll = long long;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<int> d(n), cnt(13, 0);
    cnt[0] = 1;
    for (int i = 0; i < n; i++) {
        cin >> d[i];
        cnt[d[i]]++;
    }
    vector<int> t, t2;
    for (int i = 0; i <= 12; i++) {
        if (cnt[i] >= 3) {
            cout << 0 << endl;
            return 0;
        } else if (cnt[i] == 2) {
            t2.push_back(i);
            t2.push_back(24 - i);
        } else if (cnt[i] == 1) {
            t.push_back(i);
        }
    }

    int ans = 0;
    for (int i = 0; i < (1 << t.size()); i++) {
        vector<int> t3(t2);
        for (int j = 0; j < t.size(); j++) {
            if (i & (1 << j)) t3.push_back(24 - t[j]);
            else t3.push_back(t[j]);
        }
        int hoge = 114514;
        for (int j = 0; j < t3.size(); j++) {
            for (int k = j + 1; k < t3.size(); k++) {
                int tmp = abs(t3[j] - t3[k]);
                hoge = min(hoge, min(tmp, 24 - tmp));
            }
        }
        ans = max(ans, hoge);
    }
    cout << ans << endl;
    return 0;
}