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<vector>
#include<algorithm>

using namespace std;

int main(){
	int n;
	cin >> n;
	vector<int> c(25);
	c[0] = 1;
	for(int i = 0; i < n; i++){
		int x;
		cin >> x;
		c[x]++;
	}
	c[24] = c[0];
	int p = 0;
	for(int i = 1; i <= 11; i++){
		if(c[i] >= 1){
			c[24 - i] = (c[i] + 1 - p) / 2;
			c[i] = (c[i] + p) / 2;
			p = 1 - p;
		}
	}
	int t = 1, ans = 24;
	for(int i = 1; i <= 24; i++){
		if(c[i] >= 2){
			ans = 0;
			break;
		}else if(c[i] == 1){
			ans = min(ans, t);
			t = 1;
		}else{
			t++;
		}
	}
	cout << ans << endl;
	return 0;
}
