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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using ll = long long;
using namespace std;

int main() {
    ll n;cin >> n;
    vector<int> v;
    map<int, int> mp;
    for(int i = 0;i < n;i++){
        int d;cin >> d;
        v.push_back(d);
        mp[d]++;
    }
    mp[0]++;
    vector<int> fff(12);
    int ans = 0;
    auto check = [&](){
        vector<int> vf(24);
        for(auto a : mp){
            if(a.first == 0 || a.first == 12){
                if(a.second >= 2){
                    return;
                }else{
                    vf[a.first] = 1;
                }
            }else{
                if(a.second >= 3){
                    return;
                }else if(a.second == 2){
                    vf[a.first] = 1;
                    vf[24 - a.first] = 1;
                }else{
                    if(fff[a.first] == 0){
                        vf[24 - a.first] = 1;
                    }else{
                        vf[a.first] = 1;
                    }
                }
            }
        }
        int ss = 1e9;
        for(int i = 0;i < 23;i++){
            for(int j = i + 1;j < 24;j++){
                if(vf[i] == 1 && vf[j] == 1)ss = min(j - i, ss);
            }
        }
        ans = max(ans, ss);
    };

    function<void(int)> f = [&](int i){
        if(i == fff.size()){
            check();
            return;
        }
        fff[i] = 0;f(i + 1);
        fff[i] = 1;f(i + 1);
    };
    f(0);
    check();
    cout<<ans<<endl;
}