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 <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

int main() {
    int n;
    cin >> n;

    int c[13] = { 1 };
    for (int i = 0; i < n; i++) {
        int d;
        cin >> d;

        c[d]++;
    }

    int r = 12;
    int k = 0;
    int x[24] = {};
    for (int i = 0; i <= 12; i++) {
        if (c[i] > 2 || (i % 12 == 0 && c[i] > 1)) {
            r = 0;
            break;
        }
        for (int j = 0; j < c[i]; j++) {
            x[j == k ? i : 24 - i]++;
        }
        if (c[i] == 1) k ^= 1;
    }

    int j = -r;
    for (int i = 0; i <= 24; i++) {
        if (i == 24 || x[i]) {
            r = min(r, i - j);
            j = i;
        }
    }

    cout << r << endl;

    return 0;
}