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;

#ifdef LOCAL
    #define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
        #define eprintf(...) 42
#endif

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define all(x) (x).begin(),(x).end()
#define foreach(u,v) for(auto (u) : (v))
#define pb push_back
#define mp make_pair
#define mt make_tuple

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<ll, ll> pll;
typedef vector<ll> vl;

const int inf = 1e9;
const ll linf = 1LL<<60;
const ll mod = 1e9 + 7;
const double eps = 1e-9;

/*
*/

int main()
{
  int n;
  cin >> n;
  
  vi a(n);
  rep(i, n) cin >> a[i];

  vi cnt(13);
  rep(i, n){
    cnt[a[i]]++;
    if(cnt[a[i]] >= 3){
      cout << 0 << endl;
      return 0;
    }
  }

  sort(all(a));
  
  int ans = inf;
  int l = 0, r = 24;
  rep(i, n){
    if(i%2 == 0){
      ans = min(ans, a[i]-l);
      l = a[i];
    }else{
      ans = min(ans, (24-a[i])%24);
      r = 24 - a[i];
    }
  }
  ans = min(ans, (r-l)%24);

  cout << ans << endl;
         
  return 0;
}
