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>
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define RREP(i,a,b) for(int i=(a);i>=(b);i--)
#define pq priotity_queue
#define P pair<int,int>
#define P2 pair<int,P>
#define P3 pair<int,P2>
typedef long long ll; typedef long double ld;
using namespace std;
const int INF=1e9, MOD=1e9+7, around[]={0,1,1,-1,-1,0,-1,1,0,0};
const ll LINF=1e18;
const ld PI=abs(acos(-1));
int n,a,li[30];
vector<int> vec;

int main(){
	cin >> n; li[0]=true;
	REP(i,0,n){
		cin >> a;
		if(li[a] and li[24-a]){cout << 0 << endl; return 0;}
		else if(li[a] and !li[24-a]) li[24-a]=true;
		else li[a]=true; 
	}
	
	ll ei=0;
	REP(i,0,25) if(li[i] and li[24-i]) ei+=(1<<i);
	REP(i,0,25) if(li[i]) vec.push_back(i);
	sort(vec.begin(), vec.end());
	
	int ma=0;
	REP(i,0,(1<<vec.size())){
		if(ei&i) continue;
		set<int> st;
		vector<int> pos;
		bool f=false;
		int mi=INF;
		
		REP(j,0,vec.size()){
			if((i>>j)&1 and st.count(24-vec[j])){f=true; break;}
			else if((i>>j)&1) st.insert(24-vec[j]),pos.push_back(24-vec[j]);
			else if(st.count(vec[j])){f=true; break;}
			else st.insert(vec[j]),pos.push_back(vec[j]);
		}
		
		if(f) continue;
		sort(pos.begin(), pos.end());
		REP(i,0,pos.size()-1) mi=min(mi,pos[i+1]-pos[i]);
		mi=min(mi, min(abs(pos[pos.size()-1]-pos[0]), abs(24-(pos[pos.size()-1]-pos[0]))));

		ma=max(ma,mi);
	}
		
	cout << ma << endl;
	return 0;
}
