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;

int dfs(int index,vector<int> &cnt,vector<int> &time){
    if(index==13){
        int pre=-13;
        int res=13;
        for(int i=0;i<time.size()*2;i++){
            if(time[i%24]){
                res=min(res,i-pre);
                pre=i;
            }
        }
        return res;
    }
    if(cnt[index]==0){
        return dfs(index+1,cnt,time);
    }
    if(cnt[index]==2){
        time[(12+index)%24]=true;
        time[12-index]=true;
        int res=dfs(index+1,cnt,time);
        time[(12+index)%24]=false;
        time[12-index]=false;;
        return res;
    }
    time[(12+index)%24]=true;
    int res=dfs(index+1,cnt,time);
    time[(12+index)%24]=false;
    time[12-index]=true;
    res=max(res,dfs(index+1,cnt,time));
    time[12-index]=false;
    return res;
}

int main(){
    vector<int> cnt(13,0);
    int n;
    cin>>n;
    vector<int> d(n);
    for(int i=0;i<d.size();i++){
        cin>>d[i];
        cnt[d[i]]++;
    }
    if(cnt[0]>=1 || cnt[12]>=2){
        cout<<0<<endl;
        return 0;
    }
    for(int i=1;i<cnt.size();i++){
        if(cnt[i]>=3){
            cout<<0<<endl;
            return 0;
        }
    }
    
    vector<int> time(24,0);
    time[12]=true;

    cout<<dfs(0,cnt,time)<<endl;
    return 0;
}