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;

template <typename T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;
  cin >> N;
  vector<int> D(N + 2);
  for (int i = 1; i < N + 1; i++) {
    cin >> D[i];
  }
  D[0] = 0, D[N + 1] = 24;

  int res = 0;
  for (int k = 0; k < 2 * N; k++) {
    sort(D.begin() + 1, D.end() - 1);

    int p = 0, n = 100;
    for (int i = 0; i < N + 1; i++) {
      if (chmin(n, D[i + 1] - D[i])) p = i;
    }
    if (p != 0) D[p] = 24 - D[p];

    res = max(res, n);
  }

  cout << res << '\n';
}