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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pp;
typedef pair<ll,ll> pll;
void read(int& x){ scanf("%d",&x); }
void read(ll& x){ scanf("%lld",&x); }
template<typename T,typename... Args>
void read(T& a,Args&... b){ read(a); read(b...); }
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define eb emplace_back
#define x first
#define y second

int n;

int cnt[30];

int bin;
int bi[30];
int ans;

void dfs(int p){
	if(p == bin){
		int tmp = 99;
		for(int i=0; i<24; ++i){
			for(int j=i+1; j<24; ++j){
				if(cnt[i] && cnt[j]){
					tmp = min(tmp, min(abs(i-j), 24-abs(i-j)));
				}
			}
		}
		ans = max(ans, tmp);
		return;
	}
	++cnt[bi[p]]; dfs(p+1); --cnt[bi[p]];
	++cnt[24-bi[p]]; dfs(p+1); --cnt[24-bi[p]];
}

int main()
{
	read(n);
	vector<int>v;
	for(int i=1; i<=n; ++i){
		int x; read(x);
		if(x) v.pb(x);
		else {
			puts("0");
			return 0;
		}
	}
	cnt[0]=1;
	sort(all(v));
	n = v.size();
	for(int i=0; i<n;){
		int j=i;
		while(j<n && v[i] == v[j]) ++j;
		if(j-i > 2){
			puts("0");
			return 0;
		} else if(j-i == 2){
			if(v[i] == 12){
				puts("0");
				return 0;
			}
			cnt[v[i]]++;
			cnt[24-v[i]]++;
		} else {
			bi[bin++] = v[i];
		}
		i=j;
	}
	dfs(0);
	printf("%d\n", ans);
	return 0;
}
