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 <algorithm>
#include <string>
#include <vector>
#include <functional>
#include "math.h"
using namespace std;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
#define VI vector<int>
#define debug(x) cout<<#x<<": "<<x<<endl
#define DEBUG(x) cout<<#x<<": "<<x<<endl
int N;
vector <int > D;
vector <int > num;




int main()
{

	cin >> N;

	D.resize(N);

	REP(i,N){
		cin >> D[i];
	}

	num.resize(13);
	vector <int> temp;
	REP(i,N){
		num[D[i]]++;
		if(num[D[i]]>2){
			cout << 0;
			return 0;
		}
	}

	vector <int> list;
	REP(i,13){
		if(num[i]==2){
			temp.push_back(i);
			temp.push_back(24-i);
		} else if(num[i] == 1){
			list.push_back(i);
		}
	}

	int M = list.size();
	int ans = 0;
	REP(i,pow(2,M)){
		vector <int> all = temp;
		all.push_back(0);
		REP(j,M){
			if((int)(i/pow(2,j))%2)
				all.push_back(list[j]);
			else
				all.push_back(24-list[j]);
		}

		int mi = 30;
		REP(j,all.size()-1){
			FOR(k,j+1,all.size()){
				mi = min(mi,min(max(all[j],all[k])-min(all[k],all[j]),24-max(all[j],all[k])+min(all[j],all[k])));
			}
		}
		ans = max(ans,mi);
	}
	cout << ans;
}

