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;

int n, ap[20];

int take(vector<int> v)
{
    int ans = 24;

    sort(v.begin(), v.end());

    for(int i=0; i<v.size(); ++i)
        for(int j=i+1; j<v.size(); ++j)
        {
            int a = v[i], b = v[j];

            ans = min(ans, b-a);
            ans = min(ans, 24-b+a);
        }
    return ans;
}

int main()
{
   // freopen("input", "r", stdin);
    cin.sync_with_stdio(false); cin.tie(0);

    cin >> n;

    int i, j;
    for(i=1; i<=n; ++i)
    {
        int x;
        cin >> x;
        ++ap[x];
    }

    ap[0] ++;

    vector<int> v, w;

    for(i=0; i<=12; ++i)
        if(ap[i] >= 3)
        {
            cout << 0 << '\n';
            exit(0);
        }
        else if(ap[i] == 2)
        {
            v.push_back(i);
            v.push_back(-i);
        }
        else if(ap[i] == 1)
            w.push_back(i);

    int ans = 0;

    for(i=0; i<(1<<(int)w.size()); ++i)
    {
        vector<int> vv = v;

        for(j=0; j<w.size(); ++j)
            if(i & (1<<j)) vv.push_back(w[j]);
                else vv.push_back(-w[j]);

        ans = max(ans, take(vv));
    }

    cout << ans << '\n';

    return 0;
}