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 <algorithm>
#include <iostream>
#include <string>
using namespace std;

#define REP(i, n) for (int i = 0; i < n; ++i)

int cnt[13] = {};
int place[24];

int Ans(int n) {
  int place2[24];
  copy(place, place+n, place2);
  sort(place2, place2+n);
  int ans = 12;
  REP(i, n-1) { ans = min(ans, place2[i+1] - place2[i]); }
  ans = min(ans, 24 - place2[n-1]);
  return ans;
}

int Set(int di, int i) {
  if (di == 0) {
    if (cnt[0] == 0) {
      place[0] = 0;
      return Set(1, 1);
    } else {
      return 0;
    }
  } else if (di == 12) {
    if (cnt[12] == 1) {
      place[i++] = 12;
    } else if (cnt[12] > 1) {
      return 0;
    }
    return Ans(i);
  } else {
    if (cnt[di] == 0) {
      return Set(di+1, i);
    } else if (cnt[di] == 1) {
      place[i] = di;
      int a = Set(di+1, i+1);
      place[i] = 24 - di;
      int b = Set(di+1, i+1);
      return max(a, b);
    } else if (cnt[di] == 2) {
      place[i] = di;
      place[i+1] = 24 - di;
      return Set(di+1, i+2);
    } else if (cnt[di] > 2) {
      return 0;
    }
  }
  return 0;
}

int main() {
  int n;
  cin >> n;
  REP(i, n) {
    int tmp;
    cin >> tmp;
    cnt[tmp]++;
  }

  cout << Set(0, 0) << endl;

  return 0;
}
