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 <bits/stdc++.h>
using namespace std;

int main()
{
	int N;
	cin >> N;
	if (N >= 24) {
		cout << 0 << endl;
		return 0;
	}
	vector<int> D(N), cnt(13);
	for (int i = 0; i < N; i++) {
		cin >> D[i];
		cnt[D[i]]++;
	}
	sort(D.begin(), D.end());
	if (N >= 12) {
		bool dis = cnt[0] == 0;
		for (auto val : cnt) {
			if (val > 2) {
				dis = false;
			}
		}
		cout << dis << endl;
		return 0;
	}
	int res = 0;
	for (int i = 0; i < (1 << N); i++) {
		vector<int> t(N + 1);
		for (int j = 0; j < N; j++) {
			if (i & (1 << j)) {
				t[j] = D[j];
			}
			else {
				t[j] = 24 - D[j];
			}
		}
		int ans = 24;
		for (int j = 0; j < N; j++) {
			for (int k = j + 1; k <= N; k++) {
				int d = abs(t[j] - t[k]) % 24;
				ans = min(ans, min(d, 24 - d));
			}
		}
		res = max(res, ans);
	}
	cout << res << endl;
	return 0;
}
