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;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int mod=1000000007,MAX=100001,INF=1<<30;

int main(){
    int N;cin>>N;
    vector<int> T(13,0);
    for(int i=0;i<N;i++){
        int a;cin>>a;
        T[a]++;
    }
    bool ok=true;
    if(T[0]) ok=false;
    for(int i=0;i<13;i++){
        if(T[i]>2) ok=false;
    }

    if(!ok) cout<<0<<endl;
    else{
        int ans=0;
        for(int bit=0;bit<(1<<12);bit++){
            vector<int> S(24,0);
            S[0]=1;
            for(int i=0;i<13;i++){
                if(T[i]==0) continue;
                if(T[i]==2){
                    S[i]++;
                    S[24-i]++;
                }else{
                    if(bit&(1<<(i-1))){
                        S[i]++;
                    }else S[24-i]++;
                }
            }
            int m=INF,pre=0;
            for(int i=1;i<24;i++){
                if(S[i]){
                    m=min(m,i-pre);
                    pre=i;
                }
            }
            m=min(m,24-pre);
            ans=max(ans,m);

        }

        cout<<ans<<endl;
    }
}