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>
#include <string>
#include <cmath>
#include <algorithm>
#include <utility>
#include <queue>
#include <set>
#include <map>

using namespace std;
typedef  long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
#define  MP make_pair
#define  PB push_back
#define inf  1000000007

int d[13];

int main(){
	int n;
	cin >> n;
	VI v(n);
	d[0]++;
	for(int i=0;i<n;i++){
		cin >> v[i];
		d[v[i]]++;
	}
	for(int i=0;i<13;i++){
		if(d[i]>2){
			cout << 0 << endl;
			return 0;
		}
	}
	VI a;
	int x = 0;
	for(int i=0;i<13;i++){
		if(d[i]==1){
			if(x==0){
				a.push_back(i);
				x = 1;
			}else{
				a.push_back(-i);
				x = 0;
			}
		}else if(d[i]==2){
			a.push_back(i);
			a.push_back(-i);
		}
	}
	int mi = 1000;
	for(int i=0;i<a.size();i++){
		for(int j=i+1;j<a.size();j++){
			int tmp = min(abs(a[i]-a[j]),24-abs(a[i]-a[j]));
			if(tmp < mi){
				mi  = tmp;
			}
		}
	}
	cout << mi << endl;
	return 0;
}