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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
typedef long long ll;

int main(){
    int n;
    cin >> n;
    map<int,int> m;
    int d[n];
    rep(i,n){
        cin >> d[i];
        m[d[i]]++;
    }
    if(n >= 13) {
        bool can = true;
        for(auto x : m) {
            if(x.second > 2) can = false;
        }
        cout << (can? "1" : "0") << endl;
        return 0;  
    }
    int ans = 0;
    rep(i,1<<n){
        int a[n+1] = {};
        a[n] = 0;
        rep(j,n){
            if(d[j] == 0) a[j] = 0;
            else if(i & (1<<j)) a[j] = d[j];
            else a[j] = 24-d[j];
        }
        int c = 25;
        rep(j,n+1){
            for(int k = j+1; k <= n; k++) c = min(c,abs(a[j]-a[k]));
        }
        ans = max(ans,c);
    }
    cout << ans << endl;
    return 0;
}