Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

3
7 12 8

Sample Output 1

4

For example, consider the situation where it is 7, 12 and 16 o'clock in each person's city at the moment when it is 0 o'clock in Takahashi's city. In this case, the time gap between the second and third persons' cities is 4 hours.

Sample Input 2

2
11 11

Sample Output 2

2

Sample Input 3

1
0

Sample Output 3

0

Note that Takahashi himself is also a participant.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using VI = vector<int>;
using VL = vector<ll>;
#define FOR(i,a,n) for(int (i)=(a);(i)<(n);(i)++)
#define eFOR(i,a,n) for(int (i)=(a);(i)<=(n);(i)++)
#define rFOR(i,a,n) for(int (i)=(n)-1;(i)>=(a);(i)--)
#define erFOR(i,a,n) for(int (i)=(n);(i)>=(a);(i)--)
#define SORT(i) sort((i).begin(),(i).end())
#define rSORT(i,a) sort((i).begin(),(i).end(),(a))
#define all(i) (i).begin(),(i).end()
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; }return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; }return 0; }
inline void init() { cin.tie(nullptr); cout.tie(nullptr); cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }
template<class T>inline istream& operator>>(istream& input, vector<T>& v) { for (auto& elemnt : v)input >> elemnt; return input; }

int main() {
    init();

    int n;
    cin >> n;
    VI d(13, 0);
    int in;
    FOR(i, 0, n) {
        cin >> in;
        d[in]++;
    }
    d[0]++;

    int ans = 0;
    FOR(i, 0, 1 << 11) {
        VI p;
        FOR(j, 0, d[0])p.push_back(0);
        FOR(j, 0, 11)if (d[j + 1]) {
            if (d[j + 1] == 1) {
                if (i & (1 << j))p.push_back(j + 1);
                else p.push_back(23 - j);
            }
            else {
                p.push_back(j + 1);
                FOR(k, 1, d[j + 1])p.push_back(23 - j);
            }
        }
        FOR(j, 0, d[12])p.push_back(12);

        SORT(p);
        int Min = INF, size = p.size();
        FOR(j, 0, size) {
            if (p[j] && !p[(j + 1) % size])chmin(Min, abs(p[j] - 24)); 
            else chmin(Min, abs(p[j] - p[(j + 1) % size]));
        }
        chmax(ans, Min);
        //FOR(j, 0, size)cout << p[j] << (j == size - 1 ? "\n" : " ");
    }
    cout << ans << "\n";
}