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 int long long
using namespace std;

signed main(){
  int N;cin>>N;
  std::vector<int> v(N);
  for(int i=0;i<N;i++)cin>>v[i];
  sort(v.begin(),v.end());
  int o=0;
  if(N==23){
    for(int i=0;i<N-1;i++)if(i/2+1!=v[i])o++;
    if(o==0){cout<<1<<endl;return 0;}
  }
  if(N>=23){cout<<0<<endl;return 0;}
  int ans=0;
  std::vector<int> w(N+1);
  w[0]=0;
  for(int i=0;i<pow(2,N);i++){
    int k=1;
    for(int j=0;j<N;j++){
      if(v[j]==0){w[j+1]=0;continue;}
      if(i&k)w[j+1]=v[j];
      else w[j+1]=24-v[j];
      k*=2;
    }
    sort(w.begin(),w.end());
    int fans=12;
    for(int j=0;j<N;j++)fans=min(fans,w[j+1]-w[j]);
    fans=min(fans,w[0]+24-w[N]);
    ans=max(fans,ans);
  }
  cout<<ans<<endl;
  return 0;
}
