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 m;
vector<int> d(0);

void dfs(vector<int> dd, int id, int n){
	if (id == n){
		int tempmin = 24;
		vector<int> cp(dd.size());
		cp = dd;
		for (int i = 0; i < dd.size(); i++){
			for (int j = 0; j < i; j++){
				tempmin = min(tempmin, min(24-abs(dd[i]-dd[j]), abs(dd[i]-dd[j])));
			}
		}
		m = max(m, tempmin);
	}
	else {
		dd.push_back(d[id]);
		dfs(dd, id+1, n);
		dd.pop_back();
		dd.push_back(-d[id]);
		dfs(dd, id+1, n);
		dd.pop_back();
	}
}


int main(void){
	int n, a;

	cin >> n;

	vector<int> dd(1);
	dd[0] = 0;

	for (int i = 0; i < n; i++){
		cin >> a;
		d.push_back(a);
	}

	m = 0;

	if (n >= 24){
		;
	}
	else if (n == 23){
		sort(d.begin(), d.end());
		int ok = 1;
		for (int i = 0; i < n; i++){
			if (d[i] != i/2 + 1)
				ok = 0;
		}
		m = ok;
	}
	else {
		dfs(dd, 0, n);
	}

	cout << m << endl;
}