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 <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int c[13];
bool use[24];
int main() {
  int n;
  cin >> n;
  c[0]++;
  for(int i = 0; i < n; i++) {
    int d;
    cin >> d;
    c[d]++;
  }

  int ans = 0;
  for(int i = 0; i < (1 << 11); i++) {
    //    cout <<" aaa" << endl;
    fill(use,use+24,false);
    int tmp = 30;
    for(int j = 0; j <= 12; j++) {
      if(c[j]) {
	if(j == 0 || j == 12) {
	  if(c[j] >= 2) tmp = 0;
	  use[j] = true;
	}
	else {
	  if(c[j] >= 3) tmp = 0;
	  else if(c[j] == 2) {
	    use[j] = use[24- j] = true;
	  }
	  else if(c[j] == 1) {
	    use[i & (1 << (j - 1)) ? j : 24 - j] = true;
	  }
	}
      }
    }
    for(int j = 0; j < 24; j++) {
      if(use[j]) {
	for(int k = 0; k < j; k++) {
	  if(use[k]) {
	    tmp = min(tmp,min(24 - (j - k),j - k));
	  }
	}
      }
    }
    //    cout << ans << endl;
    ans = max(ans,tmp);
  }
  cout << ans << endl;
}
	  
	
