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 main(){
    int N;
    cin >> N;
    int C[13];
    for(int i=0;i<=12;i++) C[i]=0;
    for(int i=0;i<N;i++){
        int D;
        cin >> D;
        C[D]++;
        if(D==0){
            cout << 0 << endl;
            return 0;
        }
    }
    //chokudai を12として
    bool E[50];
    for(int i=0;i<=24;i++) E[i]=false;
    E[12]=true;

    bool p=true;
    for(int i=1;i<=12;i++){
        if(C[i]>=3){
            cout << 0 << endl;
            return 0;
        }else if(C[i]==2){
            E[12+i]=true;
            E[12-i]=true;
        }else if(C[i]==1){
            if(p) E[12+i]=true;
            else E[12-i]=true;

            p=!p;
        }
    }
    if(E[0] && E[24]){
        cout << 0 << endl;
        return 0;
    }
    E[0]=(E[0] || E[24]);

    for(int i=24;i<=48;i++) E[i]=E[i-24];
    int min=100;
    int prev=-100;
    for(int i=0;i<=48;i++){
        if(E[i]){
            int d=i-prev;
            if(min>d) min=d;
            prev=i;
        }
    }

    cout << min << endl;

    return 0;
}