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<vector>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<sstream>
#include<cmath>
#include<numeric>
#include<map>
#include<stack>
#include<queue>
#include<random>
using namespace std;

int main() {

  int n; cin >> n;
  vector<int> d(n);
  for(int i=0; i<n; i++) cin >> d[i];
  sort(d.begin(), d.end());

  vector<int> a(24, 0);
  a[0]++;
  for(int i=0; i<n; i++){
    if( i % 2 == 0 ) a[ d[i]%24 ]++;
    else a[ (24-d[i])%24 ]++;
  }

  int mn = 1e9;
  for(int i=0; i<24; i++){
    if( a[i] > 1 ){
      cout << 0 << endl;
      return 0;
    }
    if( a[i] == 0 ) continue;
    for(int j=i+1; j<24; j++){
      if( a[j] == 0 ) continue;
      mn = min(mn, (j-i));
    }
  }
  cout << (mn==24 ? 0 : mn) << endl;

  return 0;
}

// EOF
