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 <iostream>
#include<bits/stdc++.h>
using namespace std;
int T[25],n,ans,rans,a;
vector<int>v;
int main(void){
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a;
        T[a]++;
        T[24-a]++;
        if(a==12)T[a]--;
    }
    if(T[0]){cout<<0<<endl;return 0;}
    for(int i=0;i<(1<<11);i++){
        v={0,24};
        for(int j=0;j<12;j++){
            if(T[j+1]>2){
                cout<<0<<endl;
                return 0;
            }
            if(T[j+1]==2){
                if(j==11){
                    cout<<0<<endl;
                    return 0;
                }
                v.push_back(j+1);
                v.push_back(23-j);
            }
            if(T[j+1]==1){
                if(((i>>j)&1)==1){
                    v.push_back(j+1);
                }
                else v.push_back(23-j);
            }
        }
        ans=30;
        sort(v.begin(),v.end());
        //for(int i=0;i<v.size();i++)cout<<v[i]<<" ";
        //cout<<endl;
        for(int i=0;i<v.size()-1;i++){
            ans=min(ans,v[i+1]-v[i]);
        }
        //cout<<ans<<endl;
        rans=max(ans,rans);
    }
    cout<<rans<<endl;
}
