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;
using ll = long long;
using db = double;
#define fi first
#define se second
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define siz(v) (ll)(v).size()
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define repi(i,x,n) for(ll i=x;i<(ll)(n);i++)
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
const ll mod = 1000000007;
const ll INF = 1000000099;
vector<ll> dx={-1,1,0,0},dy={0,0,-1,1};
//cin.tie(0);
//ios::sync_with_stdio(false);



signed main(){
    int n,ans;cin>>n;
    vector<int> d(n);
    rep(i,n)cin>>d[i];

    sort(all(d));
    ans=d[0];

    for(int i=0;i < n;i++)
    {
        if(i%2==0){
            d[i]=24-d[i];
        }
    }

    sort(all(d));
    
    

    for(int i=0;i < n-1;i++)
    {
        ans=min(ans,d[i+1]-d[i]);
    }

    cout<<ans<<endl;;


}