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<algorithm>
#include<vector>
#include<queue>
#include<string>
#include<complex>
#define lol(i,n) for(int i=0;i<n;i++)
typedef long long ll;
using namespace std;
int a[60];
int n;
bool done[60];
bool F(int lim){
	lol(i,n)done[i]=false;
	int bef=-100;
	lol(i,n){
		if(a[i]-bef<lim)continue;
		bef=a[i];
		done[i]=true;
	}
	for(int i=n-1;i>=0;i--){
		if(done[i]==false){
			int x=24-a[i];
			if(x>24-lim)return false;
			if(x-bef<lim)return false;
			bef=x;
		}
	}
	return true;
}
int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	a[n]=0;
	n++;
	sort(a,a+n);
	int maxi=-1;
	for(int i=0;i<=12;i++){
		if(F(i)==true)maxi=i;
	}
	cout<<maxi<<endl;
	return 0;
}
