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;

const double eps = 1e-10;
const int MOD = 1000000007;
const int INF = 1000000000;
const ll LINF = 1ll<<50;

template<typename T>
void printv(const vector<T>& s) {
  for(int i=0;i<(int)(s.size());++i) {
    cout << s[i];
    if(i == (int)(s.size())-1) cout << endl;
    else cout << " ";
  }
}

int main () {
  cin.tie(0);
  cout << fixed << setprecision(10);
  int n; cin >> n;
  vector<int> d(n);
  for(int i=0;i<n;++i) {
    cin >> d[i];
    d[i] = min(d[i], 24 - d[i]);
  }
  sort(d.begin(), d.end());
  // printv(d);
  vector<int> t(n+1);
  t[n] = 0;
  bool now = 0;
  for(int i=0;i<n;++i) {
    if(now) {
      t[i] = d[i];
    } else {
      t[i] = 24 - d[i];
    }
    now = !now;
  }
  
  sort(t.begin(), t.end());
  
  // printv(t1);
  // printv(t2);

  int mi = INF;

  for(int i=0;i<n+1;++i) {
    for(int j=i+1;j<n+1;++j) {
      mi = min(mi, min(abs(t[i] - t[j]), 24 - abs(t[i] - t[j])));
    }
  }

  cout << mi << endl;

}
