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>
using namespace std;

int main(){
    int n;
    cin >> n;
    vector<int> a(13, 0);   a[0]++;
    for(int i = 0; i < n; i++){
        int d;  cin >> d;
        a[d]++;
        if((n == 1 && d == 0) || a[d] > 2){
            cout << 0 << endl;
            return 0;
        }
    }
    if(a[0] == 2){
        cout << 0 << endl;
        return 0;
    }
    int ans = 0;
    for(int i = 0; i < 1<<11; i++){
        vector<int> tim;
        tim.push_back(0);
        if(a[12])   tim.push_back(12);
        for(int j = 0; j < 11; j++){
            if(a[j+1] == 0)   continue;
            if(a[j+1] == 2)   tim.push_back(j+1), tim.push_back(24-j-1);
            else if((i>>j)&1) tim.push_back(j+1);
            else              tim.push_back(24-j-1);
        }
        sort(tim.begin(), tim.end());
        int gap = 24-tim[tim.size()-1];
        for(int j = 1; j < tim.size(); j++) gap = min(gap, tim[j]-tim[j-1]);
        ans = max(ans, gap);
    }
    cout << ans << endl;
    return 0;
}