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 <algorithm>
#include <cmath>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
#define U(v) cerr << #v << ": " << (v) << endl

int main() {
    int N;
    cin >> N;
    int D[13]{};
    for (int i = 0; i < N; ++i) {
        int d;
        cin >> d;
        ++D[d];
    }
    if (D[0] || D[12] >= 2) {
        cout << 0 << endl;
        return 0;
    }
    for (int d = 1; d < 12; ++d) {
        if (D[d] >= 3) {
            cout << 0 << endl;
            return 0;
        }
    }
    bool a[24]{};
    a[0] = true;
    vector<int> b;
    for (int d = 1; d < 12; ++d) {
        if (D[d] == 2)
            a[d] = a[24 - d] = true;
        else if (D[d] == 1)
            b.emplace_back(d);
    }
    if (D[12])
        a[12] = true;
    bool c[24];
    int mx = -1 << 30;
    for (int i = 0; i < (1 << b.size()); ++i) {
        copy_n(a, 24, c);
        for (int j = 0; j < (int)b.size(); ++j) {
            if ((i >> j) & 1) {
                c[24 - b[j]] = true;
            } else {
                c[b[j]] = true;
            }
        }
        int mn = 1 << 30;
        for (int j = 0; j < 24; ++j) {
            for (int k = j + 1; k < 24; ++k) {
                if (c[j] && c[k])
                    mn = min(mn, min(k - j, j - k + 24));
            }
        }
        mx = max(mx, mn);
    }
    cout << mx << endl;
    return 0;
}
