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 <cstdio>
#include <vector>
#include <cstdlib>
#include <algorithm>

int main() {
  int N;
  scanf("%d", &N);

  if (N > 23)
    return !printf("0\n");

  std::vector<int> D(N);
  for (int i=0; i<N; ++i) {
    scanf("%d", &D[i]);
  }

  D.push_back(!++N);

  int A[13]={};
  for (int i=0; i<N; ++i)
    if (++A[std::min(D[i], 24-D[i])] > 2)
      return !printf("0\n");

  if (N == 24)
    return !printf("1\n");

  auto f=[](int d, int bits, int i)->int {
    return (bits >> i & 1)? d:24-d;
  };

  int res=0;
  for (int bits=0; bits<(1<<N); ++bits) {
    int cur=24;
    for (int i=0; i<N; ++i) {
      for (int j=i+1; j<N; ++j) {
        int a=f(D[i], bits, i);
        int b=f(D[j], bits, j);
        int d=std::abs(a-b);
        cur = std::min(cur, std::min(d, 24-d));
      }
    }
    res = std::max(res, cur);
  }

  printf("%d\n", res);
  return 0;
}
