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 FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

int cnt[13],d[51];
int main(){
    int n;
    cin>>n;
    cnt[0]=1;
    FOR(i,1,n+1){
        cin>>d[i];
        ++cnt[d[i]];
    }
    if(cnt[0]>=2||cnt[12]>=2){
        cout<<0<<endl;
        return 0;
    }
    rep(i,13){
        if(cnt[i]>=3){
            cout<<0<<endl;
            return 0;
        }
    }
    int ans=0;
    rep(i,(1<<11)){
        int mn=12;
        vector<int> v;
        v.pb(0);
        if(cnt[12]==1) v.pb(12);
        FOR(j,1,12){
            if(cnt[j]==2){
                v.pb(j);v.pb(24-j);
            }
            if(cnt[j]==1){
                if(1&(i>>(j-1))){
                    v.pb(24-j);
                }
                else v.pb(j);
            }
        }
        for(auto it1:v)for(auto it2:v){
            if(it1==it2) continue;
            int sub=min(abs(it1-it2),24-abs(it1-it2));
            mn=min(sub,mn);
        }
        ans=max(ans,mn);
    }
    cout<<ans<<endl;
    return 0;
}