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 int long long
const double PI = 3.14159265358979323846;
typedef vector<int> vint;
typedef pair<int, int> pint;
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

int N;
int D[55];
bool ex[24];

signed main() {
    cin>>N;
    for(int i=0;i<N;i++)cin>>D[i];
    sort(D,D+N);
    ex[0]=true;
    bool r=true;
    for(int i=0;i<N;i++){
        int idx;
        if(r){
            idx=D[i];
        }else{
            idx=(24-D[i])%24;
        }
        if(ex[idx]){
            cout<<0<<endl;
            return 0;
        }
        ex[idx]=true;
        r=!r;
    }
    int ans=114514,last=0;
    for(int i=1;i<48;i++){
        if(ex[i%24]){
            int x=i%24-min(last,i%24);
            last-=min(last,i%24);
            int d=abs(last-x);
            ans=min({ans,d,24-d});
            last=i%24;
        }
    }
    cout<<ans<<endl;
    return 0;
}