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 "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "iomanip"
#include "cmath"

using namespace std;

const long long int MOD = 1000000007;

long long int N, M, K, H, W, L, R;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> N;
	vector<int>v(N);
	for (int i = 0; i < N; i++) {
		cin >> v[i];
	}
	vector<int>ti(13);
	for (int i = 0; i < N; i++) {
		ti[v[i]]++;
		if (ti[v[i]] == 3) {
			cout << 0 << endl;
			return 0;
		}
	}
	if (ti[0]) {
		cout << 0 << endl;
		return 0;
	}
	if (ti[12] == 2) {
		cout << 0 << endl;
		return 0;
	}
	if (N > 13) {
		cout << 1 << endl;
		return 0;
	}
	int ans = 0;
	for (int i = 0; i < pow(2, N); i++) {
		vector<int>box;
		box.push_back(0);
		for (int j = 0; j < N; j++) {
			if (i >> j & 1) {
				box.push_back(v[j]);
			}
			else {
				box.push_back(-v[j]);
			}
		}
		sort(box.begin(), box.end());
		int bag = 100;
		for (int i = 0; i < N; i++) {
			bag = min(bag, box[i + 1] - box[i]);
		}
		bag = min(bag, box[0] + 24 - box[N]);
		ans = max(ans, bag);
	}
	cout << ans << endl;
	return 0;
}