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 <iostream>
#include <vector>
using namespace std;
int solve(int s, int N, vector<int> &D, vector<int> t, int d){
  if(s == N || d == 0) return d;
  int x = D[s], y = (24-D[s])%24, d1 = d, d2 = d;
  for(int i = 0; i <= s; ++i){
    d1 = min(d1,min(abs(t[i]-x),24-abs(t[i]-x)));
    d2 = min(d2,min(abs(t[i]-y),24-abs(t[i]-y)));
  }
  vector<int> t1 = t, t2 = t;
  t1.push_back(x);
  t2.push_back(y);
  return max(solve(s+1,N,D,t1,d1), solve(s+1,N,D,t2,d2));
}

int main(){
  int N;
  cin >> N;
  vector<int> D(N);
  for(int i = 0; i < N; ++i) cin >> D[i];
  if(N > 35){
    cout << "0" << endl;
    return 0;
  }
  vector<int> init(1,0);
  cout << solve(0,N,D,init,24) << endl;
  return 0;
}
