Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

3
7 12 8

Sample Output 1

4

For example, consider the situation where it is 7, 12 and 16 o'clock in each person's city at the moment when it is 0 o'clock in Takahashi's city. In this case, the time gap between the second and third persons' cities is 4 hours.

Sample Input 2

2
11 11

Sample Output 2

2

Sample Input 3

1
0

Sample Output 3

0

Note that Takahashi himself is also a participant.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <cassert>
#include <algorithm>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <vector>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define all(a) (a).begin(), (a).end()

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

using ll = long long;

ll N;
ll D[60];
bool dp[60][30];
std::set<ll> S, nS;

bool dfs( int idx, ll bit, ll x )
{
  if( idx == N+1 )
  {
    //rep( d, 25 ) if( bit>>d&1 )
      //std::cout << d << std::endl;

    return true;
  }

  bool fl1 = true, fl2 = true;

  rep( d, 25 ) if( bit>>d&1 )
  {
    fl1 &= llabs(D[idx-1]-d) >= x;

    fl2 &= llabs(24-D[idx-1]-d) >= x;
  }

  if( fl1 && dfs( idx+1, bit|1<<D[idx-1], x ) )
    return true;

  if( fl2 && D[idx-1] && dfs( idx+1, bit|1<<(24-D[idx-1]), x ) )
      return true;

  return false;
}

// >= x
bool C( ll x )
{
  return dfs( 1, 1, x );
}

int main()
{
  std::cin >> N;
  
  rep( i, N )
    std::cin >> D[i];

  ll ng = 0, ok = 30;

  while( ok-ng > 1 )
  {
    ll mid = (ng+ok)/2;

    (C(mid) ? ng : ok) = mid;
  }

  std::cout << ok-1 << std::endl;

  //rep( j, 25 ) 
   // std::cout << "j: " << C(j) << std::endl;

  //C(5);

  //dfs( 0,0, 4);

  //std::cout << C(2) << std::endl;

  return 0;
}