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;

int main()
{
    int n;
    scanf("%d",&n);
    vector<int> x(1);
    x.reserve(n);
    scanf("%d",&x[0]);
    const int inf=1e9;
    for(int i=1;i<n;++i)
    {
        int r;
        scanf("%d",&r);
        int s=24-r,p=inf,q=p;
        for(auto &j:x)
        {
            p=min(p,min(abs(j-r),abs(24-j+r)));
            q=min(q,min(abs(j-s),abs(24-j+s)));
        }
        if(p>q)
            x.push_back(r);
        else
            x.push_back(s);
    }
    int res=min(x[0],24-x[0]);
    for(int i=0;i<n;++i)
        for(int j=i+1;j<n;++j)
            res=min(res,min(abs(x[i]-x[j]),abs(24-x[i]+x[j])));
    printf("%d\n",res);
}