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;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
const long long mod = 1000000007;
const long long inf = 11451419198109311;
const long double pi = 3.1415926535897932384626433;
typedef pair<int, int> P;
typedef pair<int, P> PP;
struct edge { int to, cost; };
int t[13];
int w[25];
signed main() {
	int n; cin >> n;
	t[0]++;
	rep(i, n) {
		int d; cin >> d; t[d]++;
	}
	vector<int>V;
	rep(i, 13) {
		if (i==0||i==12) {
			if (t[i] == 1) {
				w[i]++;
			}
			if(t[i]>1){
				cout << 0 << endl;
				return 0;
			}
		}
		else {
			if (t[i] == 2) {
				w[i]++;
				w[24 - i]++;
			}
			else if (t[i] > 2) {
				cout << 0 << endl;
				return 0;
			}
			else if (t[i] == 1) {
				V.push_back(i);
			}
		}
	}
	int res = 0;

	for (int i = 0; i < (1 << V.size()); i++) {
		int nw[30];
		rep(j, 24)nw[j] = w[j];
		rep(j, V.size()) {
			if (i&(1 << j))nw[V[j]] = 1;
			else nw[24 - V[j]] = 1;
		}
		vector<int>ans;
		rep(j, 24) {
			if (nw[j]) {
				ans.push_back(j);
				ans.push_back(24 + j);
			}
			
		}
		sort(ans.begin(), ans.end());
		//rep(j, ans.size())cout << ans[j] << endl; cout << endl;
		int k = 24;
		rep(j, ans.size() - 1) {
			k = min(k, ans[j + 1] - ans[j]);
		}
		res = max(res, k);
	}
	cout << res << endl;
}