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;

const int INF = (int)1.01e9;

int cnt[24];

int main() {
#ifdef HOME
    freopen("in", "r", stdin);
#endif

    int n;
    while (cin >> n) {
        vector<int> d(n);
        for (int i = 0; i < n; i++) {
            cin >> d[i];
        }

        if (n > 23) {
            cout << 0 << endl;
            continue;
        }
        int ans = 0;
        for (int mask = 0; mask < (1 << n); mask++) {
            memset(cnt, 0, sizeof(cnt));
            for (int i = 0; i < n; i++) {
                if (mask & (1 << i)) {
                    if (cnt[d[i]]++ > 0) break;
                } else {
                    if (cnt[(24 - d[i]) % 24]++ > 0) break;
                }
            }
            cnt[0]++;
            int mn = INF;
            int last = -INF;
            for (int i = 0; i < 2 * 24; i++) {
                if (cnt[i % 24]) {
                    if (cnt[i % 24] > 1) mn = 0;
                    mn = min(mn, i - last);
                    last = i;
                }
                if (mn == 0) break;
            }
            ans = max(ans, mn);
            if (ans == 24 / (n + 1)) break;
        }
        cout << ans << endl;
    }


#ifdef HOME
    cerr << clock() / (double)CLOCKS_PER_SEC << endl;
#endif
    return 0;
}