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.

Examples

Input

3
7 12 8


Output

4


Input

2
11 11


Output

2


Input

1
0


Output

0
#include <bits/stdc++.h>
using namespace std;

#define fs first
#define sc second
#define pb emplace_back
#define mp make_pair
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

signed main() {
  int n;
  cin >> n;

  vector<int> d(n);
  for (auto &i : d) cin >> i;
  d.push_back(0);

  if (n >= 24) {
    cout << 0 << endl;
    return 0;
  }
  if (n >= 12) {
    int a[13] = {};
    for (int i = 0; i <= n; ++i) {
      a[d[i]]++;
    }
    // for (int i = 0; i < 13; ++i) {
    //   cout << i << " : " << a[i] << endl;
    // }

    bool f = a[0] < 2 && a[12] < 2;
    for (int i = 1; i < 12; ++i) {
      if (a[i] >= 3) {
        f = false;
      }
    }

    cout << f << endl;

    return 0;
  }

  int ans = 0;
  vector<int> a(n + 1);
  for (int i = 0; i < (1 << n); ++i) {
    for (int j = 0; j < d.size(); ++j) {
      if (i & (1 << j)) {
        a[j] = d[j];
      } else {
        a[j] = 24 - d[j];
      }
    }

    int diff = 1001001;
    for (int j = 0; j < a.size(); ++j) {
      for (int k = 0; k < a.size(); k++) {
        if (j == k) continue;
        diff = min(diff, min(abs(a[j] - a[k]), 24 - abs(a[j] - a[k])));
      }
    }

    ans = max(diff, ans);
  }

  cout << ans << endl;
}
