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;
using ll = long long;

int minD(vector<int> T, ll b) {
  for (int i = 0; i < 11; i++) {
    if (T[i + 1] and (b >> i) & 1) {
      T[i + 1]--; T[24 - (i + 1)]++;
    }
  }
  int res { 24 };
  T[24]++;
  for (int i = 0; i <= 24; i++) {
    if (T[i] == 2) return 0;
    for (int j = i + 1; j <= 24; j++) {
      if (T[i] and T[j]) res = min(res, j - i);
    }
  }
  return res;
}

int main() {
  int N; cin >> N;
  vector<int> T(25, 0);
  for (int i = 0; i < N; i++) {
    int D; cin >> D;
    T[D]++;
  }
  T[0]++;
  int ans { -1 };
  if (T[0] > 1 or T[12] > 1) ans = 0;
  for (int i = 0; i < 11; i++) {
    if (T[i + 1] > 2) ans = 0;
  }
  if (ans) {
    for (ll i = 0; i < (1 << 11); i++) ans = max(ans, minD(T, i));
  }
  cout << ans << endl;
  return 0;
}
