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>
#define REP(i,n) for(int i=0,i##_len=(n);i<i##_len;++i)
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define All(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
string binal(int k){
    string res;
    REP(i,23){
        res+=char('0'+k%2);
        k/=2;
    }
    return res;
}
int main(){
    int N;cin>>N;
    typedef pair<int,int> P;
    vector<int> D(N);
    vector<int> A(13,0);
    REP(i, N){
        cin >> D[i];
        if(D[i]==0){
            cout<<0<<endl;
            return 0;
        }
        A[D[i]]++;
        if(A[D[i]]>=3){
            cout<<0<<endl;
            return 0;
        }
    }
    int ans=23;
    vector<bool> B(25,false);
    B[24]=B[0]=true;
    bool left=true;
    REP(i,13){
        if(A[i]==2){
            B[i]=true;
            B[24-i]=true;
        }
        else if(A[i]==1){
            if(left) B[i]=true;
            else B[24-i]=true;
            left=!left;
        }
    }
    int tmp=24;
    REP(i,25){
        if(!B[i]) tmp++;
        else{
            ans=min(ans,tmp+1);
            tmp=0;
        }
    }
    cout<<ans<<endl;
}