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 <algorithm>
#include <cstring>
#include <deque>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using ll = long long;

int N;
int D[100];

int dp[100][30][30];
int solve(int i, int l, int r) {
  int &res = dp[i][l][r];
  if (res != -1) {
    return res;
  }
  if (i >= N) {
    return res = r - l;
  }
  // left
  res = max(res, min(D[i] - l, solve(i + 1, D[i], r)));
  // right
  res = max(res, min(r - (24 - D[i]), solve(i + 1, l, 24 - D[i])));
  return res;
}

int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> D[i];
  }
  sort(D, D + N);
  memset(dp, -1, sizeof(dp));
  cout << solve(0, 0, 24) << endl;
  return 0;
}
