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 <vector>
#include <cmath>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <functional>
using namespace std;

typedef long long ll;

int n;
int d[53];

int main() {
	cin >> n;
	for (int i = 0; i < n; ++i) {
		cin >> d[i];
	}
	sort(d, d + n);
	for (int i = 1; i < n; i += 2) d[i] = 24 - d[i];
	int ans = d[0];
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			if (i == j) continue;
			ans = min(ans, (int)abs(d[i] - d[j]));
		}
	}
	cout << ans << endl;
	return 0;
}