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>
#include <unordered_map>
typedef long long ll;
#define rep(i,m,n) for(ll i = (m); i < (n); i++)
#define rrep(i,m,n) for(ll i = (m); i >= (n); i--)
#define print(x) cout << (x) << endl;
#define print2(x,y) cout << (x) << " " << (y) << endl;
#define printa(x,n) for(ll i = 0; i < n; i++){ cout << (x[i]) << " \n"[i == n-1];}
#define printp(x,n) for(ll i = 0; i < n; i++){ cout << "(" << x[i].first << ", " << x[i].second << ") "; } cout << endl;
#define INF (1e18)
using namespace std;
const ll MOD = 1e9 + 7;
typedef pair<ll, ll> lpair;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N;
    ll D[55];
    cin >> N;
    rep(i,0,N) cin >> D[i];
    sort(D, D+N);
    ll lv = 0, rv = 0;
    ll pos[25] = {};
    pos[0] = 1;
    rep(i,0,N){
        if(D[i] - lv > D[i] - rv){
            pos[24-D[i]]++;
            lv = D[i];
        }else{
            pos[D[i]]++;
            rv = D[i];
        }
    }
    rep(i,0,24){
        if(pos[i] >= 2){
            print(0);
            return 0;
        }
    }
    vector<ll> tt;
    rep(i,0,24){
        if(pos[i] == 1){
            tt.push_back(i);
        }
    }
    ll ans = INF;
    ll n = tt.size();
    rep(i,0,n){
        ans = min(ans, abs(tt[(i+1)%n] - tt[i]));
    }
    print(ans);


    
}
