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 <stdio.h>
#include <algorithm>
#include <vector>
typedef long long ll;
using namespace std;

int main(void) {
  ll i, j, k, n, d, ans = 0;
  scanf("%lld", &n);
  if(n >= 24) {
    printf("0");
    return 0;
  }
  ll cnt[12];
  for(i = 0; i < 12; ++i) cnt[i] = 0;
  for(i = 0; i < n; ++i) {
    scanf("%lld", &d);
    if(!d) {
      printf("0");
      return 0; 
    }
    cnt[--d]++;
    if(cnt[d] == 3 || cnt[11] == 2) {
      printf("0");
      return 0;
    }
  }
  for(i = 0; i < (1 << 11); ++i) {
    ll min = 24;
    vector<ll> vec;
    vec.push_back(0), vec.push_back(24);
    for(j = 0; j < 12; ++j) {
      if(cnt[j] == 2) {
        vec.push_back(j + 1), vec.push_back(23 - j);
      } else if(cnt[j]) {
        if((1 << j) & i) vec.push_back(j + 1);
        else vec.push_back(23 - j);
      }
    }
    sort(vec.begin(), vec.end());
    for(j = 0; j < vec.size() - 1; ++j) if(vec[j + 1] - vec[j] < min) min = vec[j + 1] - vec[j];
    if(ans < min) ans = min;
  }
  printf("%lld", ans);
  return 0;
}