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<cstdio>
#include<cstring>
#include<cmath>
#define mid ((l+r)/2)
using namespace std;
const int maxn=50+10;
int n,d[maxn];
bool cmp(int a,int b){
	return a>b;
}
int cal(int a,int b){
	return ((a-b)%24+24)%24;
}
bool check(int c){
	int a[maxn]={0},last=d[1];
	for(int i=1;i<=n;i++)
		a[i]=d[i];
	for(int i=2;i<=n;i++)
		if(last-a[i]<c)a[i]*=-1;
		else last=a[i];
	for(int i=0;i<=n;i++)
		for(int j=0;j<=n;j++)
			if(i!=j && min(cal(a[i],a[j]),cal(a[j],a[i]))<c)
	return false;
	return true;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&d[i]);
	sort(d+1,d+n+1,cmp);
	int l=0,r=12;
	while(l+1<r){
		if(check(mid))l=mid;
		else r=mid-1;
	}
	if(check(r))printf("%d\n",r);
	else printf("%d\n",l);
	return 0;
}