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() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	if (24 <= n) {
		cout << 0 << endl;
		return 0;
	}
	int a[n];
	for (int &i : a) {
		cin >> i;
	}
	int ans = 0;
	for (int i = 0; i < (1 << n); i++) {
		int mini = numeric_limits<int>::max();
		bool puni[24]{};
		for (int j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				if (puni[a[j]]) {
					goto next;
				} else {
					puni[a[j]] = true;
				}
			} else {
				int aa = (24 - a[j]) % 24;
				if (puni[aa]) {
					goto next;
				} else {
					puni[aa] = true;
				}
			}
		}
		if (puni[0]) {
			goto next;
		}
		puni[0] = true;
		for (int i = 0; i < 24; i++) {
			for (int j = 0; j < 24; j++) {
				if (i == j) {
					continue;
				}
				if (puni[i] && puni[j]) {
					mini = min({mini, abs(i - j), 24 - abs(i - j)});
				}
			}
		}
		ans = max(ans, mini);
		next:;
	}
	cout << ans << endl;
	return 0;
}