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<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
vector<int> e;
int a[13],bes;
void dfs(int i){
	if(i == 13){
		vector<int> f(e);
		sort(f.begin(),f.end());
		f.push_back(24);
		int cur = 99;
		for(int i=0; i+1<f.size(); i++){
			int w = f[i+1]-f[i];
			cur = min(cur, min(w, 24-w));
		}
		if(cur > bes)bes = cur;
		return;
	}
	if(a[i] == 0){
		dfs(i+1);
		return;
	}
	if(i == 0 || i == 12){
		e.push_back(i);
		dfs(i+1);
		e.pop_back();
		return;
	}
	if(a[i] == 2){
		e.push_back(i);
		e.push_back(24-i);
		dfs(i+1);
		e.pop_back();
		e.pop_back();
		return;
	}
	e.push_back(i);
	dfs(i+1);
	e.pop_back();
	e.push_back(24-i);
	dfs(i+1);
	e.pop_back();
}
int main(){
	int n,x,mx = 1;
	a[0] ++;
	for(scanf("%d",&n); n--;){
		scanf("%d",&x);
		a[x] ++;
		if(a[x] > mx)mx = a[x];
	}
	if(a[0] >= 2 || a[12] >= 2 || mx >= 3){
		puts("0");
		return 0;
	}
	bes=0;
	dfs(0);
	printf("%d\n",bes);
	return 0;
}
