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;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

//#define int long long
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define rrep(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define all(c) begin(c), end(c)
const int INF =
    sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
const double PI = acos(-1);
const double EPS = 1e-9;
template <class T> bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T> bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

int dfs(vector<int> &v, vector<int> t, int index) {
  if (index == 13) {
    vector<int> x;
    rep(i, 0, t.size()) {
      if (t[i])
        x.push_back(i);
    }
    rep(i, 0, t.size()) {
      if (t[i])
        x.push_back(i);
    }
    int ret(INF);
    rep(i, 0, x.size() - 1) {
      chmin(ret, min(abs(x[i] - x[i + 1]), 24 - abs(x[i] - x[i + 1])));
    }
    return ret;
  }

  if (v[index] == 0) {
    return dfs(v, t, index + 1);
  } else if (v[index] == 1) {
    t[index]++;
    int a = dfs(v, t, index + 1);
    t[index]--;
    t[(24-index)%24]++;
    int b = dfs(v, t, index + 1);
    return max(a, b);
  }
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N;
  cin >> N;
  vector<int> D(N);
  rep(i, 0, N) cin >> D[i];
  vector<int> t(25, 0);
  t[0] = 1;

  vector<int> v(13, 0);
  rep(i, 0, D.size()) { v[D[i]]++; }
  rep(i, 0, v.size()) {
    if (v[i] >= 3) {
      cout << 0 << endl;
      return 0;
    }
  }
  rep(i, 0, v.size()) {
    if (v[i] == 2) {
      t[i]++;
      t[(24-i)%24]++;
      v[i] = 0;
    }
  }
  cout << dfs(v, t, 1) << endl;

  return 0;
}