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 <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <map>
#include <queue> 
#include <stack>
#include <set>
#include <list>

using namespace std;
typedef long long ll;
#define rep(i, e) for (int(i) = 0; (i) < (e); ++(i))
#define all(x) x.begin(),x.end()


int main()
{
	int n; cin >> n;
	vector<int> v1(25, 0);
	v1[0] = v1[24] = 1;
	rep(i, n) {
		int d; cin >> d;
		++v1[d];
	}
	if (v1[0] > 1) {
		cout << 0 << endl;
		return 0;
	}
	int res = 0;
	rep(i, pow(2, 12)) {
		vector<int> v = v1;
		rep(j, 12) {
			if (((i >> j) & 1) && v[j] > 0) {
				--v[j];
				++v[24 - j];
			}
		}
		int mn = 12;
		if (i == pow(2, 11)) {
			int a = 0;
		}
		for (int j = 1; j < 25; ++j) {
			int cnt = 0;
			if (v[j] > 1)mn = 0;
			else {
				while (j < 25 && v[j] == 0) {
					++cnt;
					++j;
				}
				if (v[j] > 1)mn = 0;
			}
			mn = min(mn, cnt+1);
		}

		res = max(res, mn);
	}
	cout << res << endl;
}