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 <cstring>
#include <iostream>
#include <map>
#include <string>
#include <vector>

using namespace std;

namespace {
#define VERA(i,a,b) for(ll i=ll(a);i<ll(b);++i)
#define ANDY(i,a,b) for(ll i=ll(a);i>ll(b);--i)
#define SIZE(x) (ll)(x.size())

typedef long long ll;
typedef double dd;

ll N;
vector<ll> D;

ll res;
vector<bool> was;

void Rec(ll per, ll low) {
  if (low <= res) return;
  if (per == N) {
    res=max(res,low);
    return;
  }
  const ll lef = D[per], rig = (24-D[per]) % 24;
  if (!was[lef]) {
    ll gap=24;
    VERA(k,0,was.size())
      if (was[k])
        gap=min(gap,min(abs(k-lef), 24-abs(k-lef)));
    was[lef]=true;
    Rec(per+1,min(low,gap));
    was[lef]=false;
  }
  if (!was[rig]) {
    ll gap=24;
    VERA(k,0,was.size())
      if (was[k])
        gap=min(gap,min(abs(k-rig), 24-abs(k-rig)));
    was[rig]=true;
    Rec(per+1,min(low,gap));
    was[rig]=false;
  }
  if (was[lef] && was[rig]) {
    res=0;
    return;
  }
}

void Solve() {
  sort(D.begin(), D.end());
  res=0;
  was.resize(24,false);
  was[0]=true;
  Rec(0,D.front());
  cout << res << endl;
}

bool Input() {
  if (!(cin>>N)) return false;
  D.resize(N);
  for(auto& d : D) cin >> d;
  return true;
}

}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  while (Input()) {
    Solve();
  }
  return 0;
}
