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 <algorithm>
#include <map>
#include <queue>
#include <bitset>
#include <set>
#include <cmath>
using namespace std;

int main(){
    int N;
    cin >> N;
    if(N >= 24){
        cout << 0 << endl;
        return 0;
    }
    vector<int> D(N);
    for(int i=0; i<N; i++)
        cin >> D[i];

    for(int i=0; i<N; i++)
        if(D[i] == 0){
            cout << 0 << endl;
            return 0;
        }

    int ans = 0;
    for(int i=0; i<(1<<N); i++){
        int vis[24];
        fill_n(vis, 24, 0);
        vis[0] = 1;
        for(int j=0; j<N; j++){
            if(i&(1<<j)) vis[D[j]]++;
            else vis[24-D[j]]++;
        }
        bool dup = false;
        for(int j=0; j<24; j++)
            if(vis[j] >= 2){
                dup = true;
                break;
            }
        if(dup) continue;

        int s = 24;
        int cnt = 24;
        for(int j=0; j<48; j++){
            cnt++;
            if(vis[j%24]){
                s = min(s, cnt);
                cnt = 0;
            }
        }
        ans = max(ans, s);
    }
    cout << ans << endl;
    return 0;
}
