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;

int main() {
    int N;
    cin >> N;
    vector<int> D(N);
    for(int i=0; i<N; ++i) cin >> D[i];
    sort(begin(D), end(D));

    int res = D[0];
    int l = D[0], r = 0;
    for(int i=1; i<N; ++i) {
        if(i % 2) {
            res = min({res, D[i] - r, 24 - l - D[i]});
            r = D[i];
        } else {
            res = min({res, D[i] - l, 24 - r - D[i]});
            l = D[i];
        }
    }
    cout << res << endl;
}