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 <iostream>
#include <algorithm>
#include <vector>
#include <math.h>
#include <bitset>
#include <tuple>
#include <time.h>
using namespace std;
typedef long long ll;


int main() {
    int n, c[13] = {}, d[55];
    
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> d[i];
        c[d[i]]++;
    }
    if (c[0] >= 1 || c[12] >= 2) {
        cout << 0 << endl;
        return 0;
    }
    int s = 0;
    for (int i = 0; i <= 12; i++) {
        if (c[i] > 2) {
            cout << 0 << endl;
            return 0;
        }
        s += c[i];
    }
    for (int i = 1; i <= 11; i++) {
        if (max(c[i], c[i + 1]) == 2 && min(c[i], c[i + 1]) == 1) {
            cout << 1 << endl;
            return 0;
        }
        s += c[i];
    }

    int ans = 0;
    vector<int> t;
    t.resize(n + 1);
    for (int i = 0; i < pow(2, n); i++) {
        t[0] = 0;
        for (int j = 0; j < n; j++) {
            if (i & (1 << j)) t[j + 1] = d[j];
            else t[j + 1] = 24 - d[j];
        }
        sort(t.begin(), t.end());
        //for (int j = 0; j < n + 1; j++) cout << t[j] << " ";
        //cout << endl;

        int mn = 24;
        for (int j = 0; j < n; j++) {
            mn = min(mn, t[j + 1] - t[j]);
        }
        mn = min(mn, 24 + t[0] - t[n]);
        ans = max(ans, mn);
    }

    cout << ans << endl;
    
}
