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;
#define rep(i, a, b) for (int i = (a); i < (b); i++)

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

    int n, s = 0;
    cin >> n;
    vector<int> d(n);
    rep(i, 0, n) cin >> d[i];
    vector<int> e(13);
    rep(i, 0, n) e[d[i]]++;

    rep(i, 0, 13) if (e[i]) {
        s = i;
        break;
    }
    rep(i, 1, 13) if (e[i] > 2 - (i / 12)) s = 0;
    if (s == 0) {
        cout << 0 << endl;
        return 0;
    }

    set<int> st;
    vector<int> t;
    st.insert(0);
    st.insert(24);

    rep(i, 1, 13) {
        if (e[i] == 2 - (i / 12)) {
            st.insert(i);
            st.insert(24 - i);
        } else if (e[i] == 1) {
            t.push_back(i);
        }
    }

    int m = t.size();
    s = 0;

    rep(i, 0, 1 << m) {
        set<int> sx = st;
        rep(j, 0, m) if (i & (1 << j)) { sx.insert(t[j]); }
        else {
            sx.insert(24 - t[j]);
        }
        int k = 24, l = -24;
        for (auto x : sx) {
            k = min(k, x - l);
            l = x;
        }
        s = max(s, k);
    }
    cout << s << endl;

    return 0;
}
