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: Time Limit Exceeded)

#include<iostream>
#include<vector>
#include<string>
#include<cmath>
#include<algorithm>
#include <numeric>
#include<map>
#include<unordered_map>
#include <queue>
 
using namespace std;
using ll=long long;
#define rep(i,n)  for(ll i=0;i<n;++i)
#define all_map(itr,mp) for(auto itr=mp.begin();itr!=mp.end();itr++)
#define ALL(a) (a).begin(),(a).end()

int main(){
    ll n;
    cin >> n;
    vector<ll> d;
    int used[24]={};
    rep(i, n){
        int in;
        cin >> in;
        if(used[in]<2)d.push_back(in);
        // used[in]++;
    }
    ll ans = 0;
    sort(ALL(d));
    ll mn = min(d.front(), 24-d.back());
    ll time = 0;
    rep(i, (1<<d.size())){
        ll s  = mn;
        rep(j, d.size()){
            ll city1;
            if(i&(1<<j))city1 = d[j];
            else city1 = 24-d[j];
            for(ll k=j+1; k<d.size(); ++k){
                ll city2;
                if(i&(1<<k))city2 = d[k];
                else city2 = 24-d[k];
                ll dif = abs(city1-city2);
                s = min(s, min(dif, 24-dif));
            }
        }
        ans = max(s, ans);
    }
    cout << ans << endl;
}