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>
#define FOR(i, a, b) for(int i=(int)(a); i!=(int)(b); i++)
#define RFOR(i, a, b) for(int i=(int)(a); i!=(int)(b); i--)
#define REP(i, n) FOR(i, 0, n)
using namespace std;
using ll = long long;

int dis(int a, int b){
  return min(abs(a-b), 24-abs(a-b));
}

int main(){
  int n; cin >> n;
  int d[n], ci[13]{}, cl[24]{};
  ci[0] = 1;
  REP(i, n){
    cin >> d[i];
    ci[d[i]]++;
  }

  int ans = 0;
  REP(i, 1<<13){
    bitset<13> b(i);

    //2^11のパターンを構築
    memset(cl, 0, sizeof(cl));
    REP(j, 13){
      if(ci[j] == 1){
        if(b[j]) cl[j] = 1;
        else cl[(24-j)%24] = 1;
      }
      else if(ci[j] == 2){
        if(j == 0 || j == 12){
          cout << 0 << endl;
          return 0;
        }
        cl[j] = cl[(24-j)%24] = 1;
      }
      else if(ci[j] > 2){
        cout << 0 << endl;
        return 0;
      }
    }

    //実際に最小値を計算
    int t = 24;
    REP(j, 24)REP(k, 24){
      if(j != k && cl[j] && cl[k]){
        t = min(t, dis(j, k));
      }
    }

    ans = max(t, ans);
  }

  cout << ans << endl;

  return 0;
}
