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 ull = unsigned long long;
const ll INF = 4e18;
const ll MOD = 1e9+7;

int a[13] = {};
bool used[24] = {};

int f(int n) {
  if (n == 13) {
    int ans = 24;
    int cnt = 1;
    for (int i = 1; i <= 24; i++) {
      if (used[i%24] == 0) cnt++;
      else {
        ans = min(ans, cnt);
        cnt = 1;
      }
    }
    return ans;
  }

  if (n == 0)
    if (a[0] != 0) return 0;
    else {used[0] = 1; return f(n+1);}
  if (n == 12)
    if (a[12] > 1) return 0;
    else {used[12] = a[12]; return f(n+1);}

  if (a[n] >= 3) return 0;
  else if (a[n] == 2) {
    used[n] = used[24-n] = 1;
    return f(n+1);
  } else if (a[n] == 1) {
    int ans = 0;
    used[n] = 1;
    ans = max(ans, f(n+1));
    used[n] = 0;
    used[24-n] = 1;
    ans = max(ans, f(n+1));
    used[24-n] = 0;
    return ans;
  } else {
    return f(n+1);
  }
}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int d;
    cin >> d;
    a[d]++;
  }
  cout << f(0) << endl;
  return 0;
}
