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: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;
const int maxn = 5050;
int mark[maxn];
int main(){
    int n;
    cin >> n;
    while(n--){
        int x; cin >> x;
        mark[x]++;
    }
    for(int e = 0; e < 13; e++) mark[e] = min(mark[e], 2);
    vector<int> val;
    for(int e = 0; e < 13; e++) while(mark[e]--) val.push_back(e);
    int ans = 0;
    // for(int e : val) cerr << e << " "; cerr << endl;
    for(int e = 0; e < 1<<val.size(); e++){
        vector<int> cloc;
        for(int f = 0; f < val.size(); f++){
            if(e&(1<<f)) cloc.push_back(24 - val[f]);
            else cloc.push_back(val[f]);
        }
        sort(cloc.begin(), cloc.end());
        // for(int f : cloc) cerr << f << " "; cerr << endl;
        int mn = 25;
        for(int f = 1; f < cloc.size(); f++) mn = min(mn, cloc[f] - cloc[f-1]);
        if(mn == 25) mn = 0;
        ans = max(ans, mn);
        // cerr << mn << endl;
    }
    cout << ans << endl;
    return 0;
}