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 <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
#include <deque>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N; cin >> N;
  int cnt[13]; CLR(cnt);
  cnt[0] = 1;
  FOR(i,0,N) {
    int d; cin >> d;
    cnt[min(d, 24-d)]++;
  }
  if(cnt[0] >= 2 || cnt[12] >= 2) {
    cout << 0 << endl;
    return 0;
  }
  FOR(i,1,12) {
    if(cnt[i] >= 3) {
      cout << 0 << endl;
      return 0;
    }
  }
  int ans = 0;
  FOR(i,0,1<<11) {
    vector<int> v;
    v.push_back(0);
    if(cnt[12]) v.push_back(12);
    FOR(j,0,11) {
      // 1~11
      if(cnt[j+1] == 1) {
        if((i>>j)&1) v.push_back(j+1);
        else v.push_back(23-j);
      }
      else if(cnt[j+1] == 2) {
        v.push_back(j+1);
        v.push_back(23-j);
      }
    }
    sort(v.begin(), v.end());
    int mn = 1e9;
    FOR(i,0,v.size()) {
      //cout << v[i] << endl;
      int j = (i + 1) % v.size();
      mn = min(mn, min(abs(v[i]-v[j]), 24-abs(v[i]-v[j])));
    }
    ans = max(ans, mn);
  }
  cout << ans << endl;
  return 0;
}