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 ll = long long;
using namespace std;
const ll INF = 1000000000000000000;
const ll mod = 1000000007;
ll test[12345678], res[12345678];
signed main() {
	vector<char> v; map<ll, ll> cost;
	ll n, m, a, b, c, x, cnt = 0, ans = 1000000; string s, t;
	cin >> n; test[0]++;
	for (int h = 0; h < n; h++) {
		cin >> a; a = min(a, 24 - a); test[h] = a;
	}
	test[n] = 0; sort(test, test + n + 1);
	for (int h = 0; h <= n; h++) {
		if (h == 0) { ans = min(ans, abs(test[h] - test[n])); }
		else { ans = min(ans, abs(test[h - 1] - test[h])); }
	}
	for (int i = 0; i <= (1 << 11); i++) {
		vector<ll> sep = { }; cnt = 10000;
		for (int h = 0; h <= n; h++) {
			res[h] = test[h];
		}
		for (int j = 0; j <= 10; j++) {
			if ((i & (1 << j)) > 0) {
				sep.push_back(j + 1);
			}
		}
		//if (sep.size() == 0) { continue; }//?
		for (int h = 0; h < sep.size(); h++) {
			for (int j = 0; j <= n; j++) {
				if (res[j] == sep[h]) { res[j] = 24 - res[j]; break; }
			}
		}
		sort(res, res + 1 + n);
		for (int h = 0; h <= n; h++) {
			//cout << i << ' ' << res[h] << endl;
			if (h == 0) { cnt = min(cnt, min(abs(res[h] - res[n]), 24 - abs(res[h] - res[n]))); }
			else { cnt = min(cnt, min(abs(res[h - 1] - res[h]), 24 - abs(res[h - 1] - res[h]))); }
		}
		ans = max(ans, cnt);
	}
	cout << ans << endl;
	return 0;
}