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, d[50];
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> d[i];
	}
	sort(d, d + n);
	int u = 0, v = 0, r = 12;
	for (int i = 0; i < n - 1; i++) {
		if (u <= v) {
			r = min(r, d[i] - u);
			u = d[i];
		} else {
			r = min(r, d[i] - v);
			v = d[i];
		}
	}
	r = min(r, max(min(d[n - 1] - u, 24 - v - d[n - 1]), min(d[n - 1] - v, 24 - u - d[n - 1])));
	cout << r << endl;
	return 0;
}
