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 <vector>
#include <limits>

using namespace std;

int main(){
    int N; cin >> N;
    vector<int> D(N);
    for(int i=0;i<N;i++) cin >> D[i];

    vector<int> cnt(13,0);
    cnt[0] += 1;
    for(int i=0;i<N;i++) cnt[D[i]]++;

    int ans = -1;

    for(int i=0;i<(1<<11);i++){
        vector<bool> used(24,false);
        int s = numeric_limits<int>::max();

        for(int j=0;j<=12;j++){
            if(cnt[j]){
                if(j==0 || j==12){
                    used[j] = true;
                    if(cnt[j]>=2){
                        s = 0;
                    }
                }else{
                    if(cnt[j] == 1){
                        used[((i & (1<<(j-1))) ? j : 24-j)] = true;
                    }else if(cnt[j] == 2){
                        used[j] = used[24-j] = true;
                    }else{
                        s = 0;
                    }
                }
            }
        }
        for(int j=0;j<24;j++){
            if(used[j]){
                for(int k=0;k<j;k++){
                    if(used[k]){
                        s = min(s, min(abs(k-j), 24 - abs(k-j)));
                    }
                }
            }
        }
        ans = max(ans, s);
    }

    cout << ans << endl;
    return 0;
}