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 <cassert>
#include <cstring>
#include <iostream>
#include <vector>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define TRACE(x) cout << #x << " = " << x << endl
#define _ << " _ " <<

typedef long long llint;

const int MAX = 55;

vector<int> E[MAX];

bool bio[MAX];
int color[MAX];

bool dfs(int x, int col) {
  if (bio[x]) return color[x] == col;
  bio[x] = true;
  color[x] = col;
  for (int y: E[x]) {
    if (!dfs(y, col ^ 1)) return false;
  }
  return true;
}

int main(void) {
  int N;
  scanf("%d", &N);
  vector<int> d(N);
  REP(i, N) scanf("%d", &d[i]);

  auto can = [&] (int s) {
    REP(i, N) E[i].clear();
    REP(i, N) REP(j, i) {
      if (abs(d[i] - d[j]) < s) {
        int d2 = min(d[i] + d[j], 24 - (d[i] + d[j]));
        if (d2 < s) return false;
        E[i].push_back(j);
        E[j].push_back(i);
      }
    }

    REP(i, N) color[i] = 0, bio[i] = false;
    REP(i, N) if (!bio[i]) {
      if (!dfs(i, 0)) return false;
    }
    return true;
  };
  
  int ret = *min_element(d.begin(), d.end());
  while (!can(ret)) ret--;
  
  printf("%d\n", ret);
  return 0;
}
