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;
 
using ll = long long;
#define debug(x) cerr << #x << ": " << x << endl;
 
int main() {
   int n;
   cin >> n;
 
   vector<int> v(n);
   for (int i = 0; i < n; i++) {
      cin >> v[i];
   }
 
   if (n == 1) {
      cout << v[0] << endl;
      return 0;
   }
 
   int Min = 10000;
   for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
         if (i == j) continue;
         int dif = abs(v[i] - v[j]);
         int dif2 = abs(24 - v[i] - v[j]);
         int tmp = max(dif, dif2);
         Min = min(dif2, Min);
      }
   }
 
   for (int i = 0; i < n; i++) {
      Min = min(Min, v[i]);
   }
 
   cout << Min << endl;
 
   return 0;
}