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 int long long
using namespace std;

signed main()
{
    int N; cin>>N;
    vector<int> cnt(13);
    cnt[0]++;
    for(int i=0;i<N;i++){
	int d; cin>>d;
	cnt[d]++;
    }

    if(cnt[0]>1 or cnt[12]>1){
	cout<<0<<endl;
	return 0;
    }
    for(int i=1;i<=11;i++){
	if(cnt[i]>2){
	    cout<<0<<endl;
	    return 0;
	}
    }

    int ans=0;
    const int K=11;
    for(int S=0;S<(1<<K);S++){
	vector<int> v={0,24};
	for(int i=0;i<K;i++){
	    if(cnt[i+1]==2){
		v.push_back(i+1);
		v.push_back(24-(i+1));
	    }
	    if(cnt[i+1]==1){
		if(S>>i&1) v.push_back(i+1);
		else v.push_back(24-(i+1));
	    }
	}
	if(cnt[12]) v.push_back(12);
	sort(v.begin(),v.end());
/*
	for(int i=0;i<v.size();i++) cout<<v[i]<<' ';
	cout<<endl;
*/
	int min_diff=24;
	for(int i=0;i<v.size()-1;i++){
	    min_diff=min(min_diff,v[i+1]-v[i]);
	}
	ans=max(ans,min_diff);
    }
    cout<<ans<<endl;
}
