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 <bitset>
#include <cctype>
#include <chrono>
#include <cmath>
#include <cstdio>
#include <ctime>
#include <deque>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <string>
#include <tuple>
#include <unordered_map>
#include <vector>
using namespace std;

#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()

const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
/*----------------------------------------*/

int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  // freopen("input.txt", "r", stdin);

  int n; cin >> n;
  vector<int> jikoku{0, 24}, d(n);
  REP(i, n) {
    cin >> d[i];
    if (d[i] == 0) {
      cout << 0 << '\n';
      return 0;
    }
  }
  sort(ALL(d));
  REP(i, n) {
    int index1 = distance(jikoku.begin(), lower_bound(ALL(jikoku), d[i])), index2 = distance(jikoku.begin(), lower_bound(ALL(jikoku), 24-d[i]));
    int sa1 = min(jikoku[index1]-d[i], d[i]-jikoku[index1-1]), sa2 = min(jikoku[index2]-(24-d[i]), (24-d[i])-jikoku[index2-1]);
    if (sa1 >= sa2) jikoku.push_back(d[i]);
    else jikoku.push_back(24-d[i]);
    sort(ALL(jikoku));
  }
  int ans = 12;
  REP(i, jikoku.size()-1) ans = min(ans, jikoku[i+1]-jikoku[i]);
  cout << ans << '\n';
  return 0;
}
