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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x)+modulo)%modulo)
#define Inf 1000000000000000000

int main(){
	
	int N;
	cin>>N;
	
	vector<int> D(N+1);
	D[N]=0;
	for(int i=0;i<N;i++){
		cin>>D[i];
	}
	
	map<int,int> mp;
	for(int i=0;i<=N;i++){
		mp[D[i]]++;
	}
	
	if(mp[0]>=2){
		cout<<0<<endl;
		return 0;
	}
	if(mp[12]>=2){
		cout<<0<<endl;
		return 0;
	}
	
	vector<int> A;
	for(int i=0;i<=12;i++){
		if(mp[i]>=3){
			cout<<0<<endl;
			return 0;
		}
		for(int j=0;j<mp[i];j++){
			A.push_back(i);
		}
	}
	if(A.size()>=13){
		cout<<1<<endl;
		return 0;
	}
	int ans = 0;
	for(int i=0;i<(1<<A.size());i++){
		vector<int> B(A.size());
		for(int j=0;j<A.size();j++){
			if((i>>j)&1){
				B[j] = A[j];
			}
			else{
				B[j] = 24-A[j];
			}
		}
		sort(B.begin(),B.end());
		int k = 99;
		for(int j=0;j<B.size();j++){
			if(j==0){
				k = min(k,24-B[B.size()-1]);
			}
			else{
				k = min(k,B[j]-B[j-1]);
			}
		}
		ans = max(k,ans);
	}
	
	cout<<ans<<endl;
	
	
	
	
	return 0;
}
