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<sstream>
#include<fstream>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#include<functional>
#include<utility>
#include<bitset>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cstdio>

using namespace std;

#define REP(i,n) for(int i=0;i<int(n);i++)
#define foreach(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
typedef long long ll;
typedef pair<int,int> P;


int main(void){
	          int i;

	      	  cin.tie(0);  ios_base::sync_with_stdio(false);

	      	  int N;
	      	  cin >> N;

	      	  vector<int> d(N+1),e(N+1);
	      	  d[0]=0; e[0]=0;
	      	  for(i=1;i<=N;i++){
	      		  cin >> d[i];
	      		  e[i]=d[i];
	      	  }

	      	  sort(d.begin(),d.end());
	      	 sort(e.begin(),e.end());


	      	  for(i=1;i<=N;i++){
	      		  if(i%2==1){
	      			  d[i]=min(d[i],24-d[i]);
	      		  }else{
	      			  d[i]=max(d[i],24-d[i]);
	      		  }
	      	  }

	      	  int mn=1e9;
	      	  for(i=1;i<=N;i++){
	      		  int x=abs(d[i]-d[i-1]);
	      		  if(i==1) mn=d[i];
	      		  if(i>=2) mn=min(min(mn,abs(d[i]-d[i-2])),min(x,24-x));
	      	  }




	      	  cout << mn << endl;



	          return 0;
}
