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 <bits/stdc++.h>
using namespace std;
#define MOD 1000000007
//#define MOD 998244353
#define INF 1145141919810893364
//#define INF 810114514
#define PI 3.141592653589
typedef pair<int,int> PP;
typedef long long ll;
#define int ll
#define setdouble setprecision
#define REP(i,n) for(int i=0;i<(n);++i)
#define OREP(i,n) for(int i=1;i<=(n);++i)
#define RREP(i,n) for(int i=(n)-1;i>=0;--i)
#define GOODBYE do { cout << "0" << endl; return 0; } while (false)
#define MM <<" "<<
#define Endl endl

signed main(void){
    int N;
    int D[57];
    cin >> N;
    D[0]=0;
    REP(i,N){
        cin >> D[i+1];
    }
    int Ans=INF;
    sort(D,D+N+1);
    REP(i,N+1){
        if(i%2==1)D[i]=24-D[i];
        if(D[i]==24)D[i]=0;
        //cout << D[i] << endl;
    }
    REP(i,N+1){
        REP(j,N+1){
            if(i==j)continue;
            int d=abs(D[i]-D[j]);
            Ans=min(Ans,d);
            Ans=min(Ans,24-d);
        }
    }
    cout << Ans << endl;
    return 0;
}

