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 <algorithm>
#include <vector>
using namespace std;

int N;
vector<int> country_time_differences{0};
vector<int> country_times;

int choose(int now){
    if(now == N + 1){
        vector<int> country_times_sorted = country_times;
        sort(country_times_sorted.begin(), country_times_sorted.end());
        int ans = country_times_sorted[0] - country_times_sorted[N] + 24;
        for(int i = 0; i < N; i++){
            ans = min(ans, country_times_sorted[i + 1] - country_times_sorted[i]);
        }
        return ans;
    }else if(country_time_differences[now] % 12 == 0){
        country_times.push_back(country_time_differences[now]);
        int ans = choose(now + 1);
        country_times.pop_back();
        return ans;
    }else{
        country_times.push_back(country_time_differences[now]);
        int ans = choose(now + 1);
        country_times.back() = 24 - country_time_differences[now];
        ans = max(ans, choose(now + 1));
        country_times.pop_back();
        return ans;
    }
}

int main(){
    cin >> N;
    for(int i = 0; i < N; i++){
        int input;
        cin >> input;
        country_time_differences.push_back(input);
    }
    if(N >= 24){
        cout << 0 << endl;
        return 0;
    }
    cout << choose(0) << endl;
}
