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 <algorithm>
#include <string>
using namespace std;

int main(){
	int n;
	cin >> n;
	vector<int> d(n), c(13,0);
	for(int i=0; i<n; i++){
		cin >> d[i];
		c[d[i]]++;
	}
	c[0]++;
	
	vector<bool> con(25, false);
	vector<int> nd;
	for(int i=0; i<13; i++){
		if(( (i==0||i==12)&&c[i]>1 ) || c[i]>2){
			cout << 0 << endl;
			return 0;
		}
		if(c[i]==2 || (c[i]==1&&(i==0||i==12))){
			con[i] = con[24-i] = true;
			c[i] = 0;
		}
		if(c[i]==1){
			nd.push_back(i);
		}
	}
	
	n = nd.size();
	int ans=0;
	if(n==0){
		int tmin = 100;
		int tmp=100;
		for(int j=0; j<24; j++){
			if(con[j]){
				tmin = min(tmin, tmp);
				tmp=1;
			}else{
				tmp++;
			}
		}
		ans = max(ans, tmin);
	}
	for(int i=0; i<(1<<n); i++){
		vector<bool> tcon = con;
		for(int j=0; j<n; j++){
			if((i&(1<<j))!=0){
				tcon[nd[j]] = true;
			}else{
				tcon[24-nd[j]] = true;
			}
		}
		int tmin = 100;
		int tmp=100;
		for(int j=0; j<25; j++){
			if(tcon[j]){
				tmin = min(tmin, tmp);
				tmp=1;
			}else{
				tmp++;
			}
		}
		ans = max(ans, tmin);
	}
	cout << ans << endl;

	return 0;
}