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;
const int kMaxValue = 12;

int main() {
    int n; cin >> n;
    vector<int> v(n, 0);
    vector<int> f(kMaxValue + 1, 0);

    for (int i = 0; i < n; i += 1) {
        cin >> v[i];
        f[v[i]] += 1;

        if (f[v[i]] >= 3) {
            cout << "0\n";
            return 0;
        }
    }

    if (f[0] >= 1) {
        cout << "0\n";
        return 0;
    }

    if (f[12] >= 2) {
        cout << "0\n";
        return 0;
    }
   
    vector<int> state (2 * kMaxValue + 1, 0); 
    vector<int> others;
    state[0] = 1;

    for (int i = 1; i <= 12; i += 1) {
        if (f[i] == 2) {
            state[i] = 1;
            state[24 - i] = 1;
        } else if(f[i] == 1) {
            others.push_back(i);
        }
    }
    
    int k = others.size();
    int ans = 0;
    
    auto eval = [] (vector<int> state) {
        int ans = kMaxValue + 1;
        for (int i = 0; i <= 23; i += 1)
            for (int j = i + 1; j <= 23; j += 1)
                if (state[i] and state[j])
                    ans = min(ans, min(j - i, 24 - (j - i)));

        return ans;
    };

    for (int mask = 0; mask < (1 << k); mask += 1) {
        vector<int> temp_state = state;
        for (int i = 0; i < k; i += 1) {
            if ((1 << i) & mask) {
                temp_state[others[i]] = 1;
            } else {
                temp_state[24 - others[i]] = 1;
            }
        }

        ans = max(ans, eval(temp_state));
    }
    
    cout << ans << "\n";
}