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 <utility>
#include <string>
#define llint long long

using namespace std;
typedef pair<llint, llint> P;
typedef pair<llint, P> E;

llint n;
llint cnt[13];
bool used[25];

int main(void)
{
	cin >> n;
	
	cnt[0] = 1;
	llint x;
	for(int i = 1; i <= n; i++){
		cin >> x;
		cnt[x]++;
	}
	for(int i = 0; i <= 12; i++){
		if(cnt[i] >= 3){
			cout << 0 << endl;
			return 0;
		}
	}
	if(cnt[0] >= 2 || cnt[12] >= 2){
		cout << 0 << endl;
		return 0;
	}
	
	int N = 1<<13, ans = 0;
	for(int i = 0; i < N; i++){
		for(int j = 0; j < 24; j++) used[j] = false;
		used[24] = true;
		for(int j = 1; j <= 12; j++){
			if(cnt[j] == 0) continue;
			if(cnt[j] == 2) used[j] = used[24-j] = true;
			if(cnt[j] == 1){
				if(i & (1<<j)) used[j] = true;
				else used[24-j] = true;
			}
		}
		//for(int j = 0; j <= 24; j++) cout << used[j] ; cout << endl;
		int cnt = 0, mn = 25;
		for(int j = 0; j <= 24; j++){
			if(used[j]) mn = min(mn, cnt), cnt = 0;
			cnt++;
		}
		ans = max(ans, mn);
	}
	cout << ans << endl;
	
	return 0;
}