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;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef pair<ll, pair<ll, ll> > P3;

const ll MOD = ll(1e9 + 7);
const ll LLINF = LLONG_MAX;
const int IINF = INT_MAX;
const int MAX_N = int(2e3) + 5;
const double EPS = 1e-8;
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
#define SORT(v) sort((v).begin(), (v).end())
#define SORTR(v) sort((v).rbegin(), (v).rend())
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)


int main() {
    int N;
    vector<int> D;
    cin >> N;
    REP(i,N){
        int d;
        cin >> d;
        D.push_back(d);
    }
    D.push_back(0);
    D.push_back(24);
    SORT(D);
    REP(s,24){
        vector<int> t = D;
        int i=1;
        while(i<t.size()){
            int dif = abs(t[i]-t[i-1]);
            if(min(dif,24-dif)>=s){
                i++;
            }
            else{
                if(t[i]<12){
                    t.push_back(24-t[i]);
                    t.erase(t.begin()+i);
                    SORT(t);
                }
                else{
                    cout << s-1 << endl;
                    return 0;
                }
            }
        }
    }
    return 0;
}
