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 fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
const LL mod=1000000007;
const LL LINF=1LL<<60;
const int INF=1<<30;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};


int n;
vector<int> s(50);
vector<int> a(52);

int dfs(int k){
    if(k==n){
        int ret = 24;
        for (int i = 0; i < n; i++) {
            for (int j = i+1; j < n; j++) {
                ret = min(ret, abs(s[i] - s[j]));
            }
        }
        return ret;
    }
    int ret = 0;
    if(a[k]==0||a[k]==12){
        s[k] = a[k];
        ret = max(ret, dfs(k+1));
    }
    else{
        s[k] = a[k];
        ret = max(ret, dfs(k+1));
        s[k] = 24 - a[k];
        ret = max(ret, dfs(k+1));
    }
    return ret;
}

 

int main(){
    cin >> n;
    a[0] = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i+1];
    }
    n++;
    if(n>24){
        puts("0");
        return 0;
    }
    cout << dfs(0) << endl;
    return 0;
}