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 cnt[15];
int main() {
	int n;
	scanf("%d", &n);
	cnt[0] = 1;
	for (int i = 0; i < n; ++ i) {
		int x;
		scanf("%d", &x);
		cnt[x] ++;
	}
	vector<int> a, c;
	for (int i = 0; i <= 12; ++ i) {
		if (cnt[i] >= 3) return 0 * puts("0");
		if (cnt[i] == 2) {
			c.push_back(i);
			c.push_back(24 - i);
		}
		else if (cnt[i]) {
			a.push_back(i);
		}
	}
	int ans = 0;
	for (int mask = 0; mask < (1 << (int) a.size()); ++ mask) {
		vector<int> b = c;
		for (int i = 0; i < (int) a.size(); ++ i) {
			b.push_back((mask >> i & 1) ? a[i] : 24 - a[i]);
		}
		int s = 100;
		for (int i = 0; i < (int) b.size(); ++ i) {
			for (int j = i + 1; j < (int) b.size(); ++ j) {
				int t = abs(b[i] - b[j]);
				s = min(s, min(t, 24 - t));
			}
		}
		ans = max(ans, s);
	}
	printf("%d\n", ans);
}
