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 "bits/stdc++.h"
#include <sys/timeb.h>
#include <fstream>

using namespace std;

#define repr(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repr(i,0,n)
#define reprrev(i,a,b) for(int i=(int)(b)-1;i>=(int)(a);i--)
#define reprev(i,n) reprrev(i,0,n)
#define repi(itr,ds) for(auto itr=ds.begin();itr!=ds.end();itr++)
#define all(a) a.begin(),a.end()
#define mp make_pair
#define mt make_tuple
#define INF 2000000000
#define INFL 1000000000000000000LL
#define EPS (1e-8)
#define MOD 1000000007
#define PI 3.141592653589793
#define RMAX 4294967295
#define equals(a,b)(fabs((a)-(b))<EPS)

typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<P> vP;
typedef vector<vector<int> > vvi;
typedef vector<vector<bool> > vvb;
typedef vector<vector<ll> > vvll;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<double> > vvd;
typedef vector<vector<P> > vvP;
typedef complex<double> comp;
typedef vector<complex<double>> vcomp;
typedef priority_queue<int, vector<int>, greater<int> > pqli;
typedef priority_queue<ll, vector<ll>, greater<ll> > pqlll;
typedef priority_queue<P, vector<P>, greater<P> > pqlP;
typedef string::const_iterator State;
class ParseError {};
struct Edge {
	int from, to, cost;
	bool operator<(Edge e) {
		return cost < e.cost;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;

class UnionFind {
public:
	vector<int> par, rank, _size;
	int _num;

	UnionFind(int n) {
		par = vector<int>(n);
		rank = vector<int>(n);
		_size = vector<int>(n);
		_num = n;
		for (int i = 0; i < n; i++) {
			par[i] = i;
			rank[i] = 0;
			_size[i] = 1;
		}
	}

	// 木の根を求める
	int find(int x) {
		if (par[x] == x) {
			return x;
		}
		else {
			return par[x] = find(par[x]);
		}
	}

	// xとyの属する集合を併合
	void unite(int x, int y) {
		x = find(x);
		y = find(y);
		if (x == y)return;

		_num--;
		if (rank[x] < rank[y]) {
			par[x] = y;
			_size[y] += _size[x];
		}
		else {
			par[y] = x;
			_size[x] += _size[y];
			if (rank[x] == rank[y])rank[x]++;
		}
	}

	// xとyが同じ集合に属するか否か
	bool same(int x, int y) {
		return find(x) == find(y);
	}

	// xの属する集合のサイズ
	int size(int x) {
		return _size[find(x)];
	}

	// 集合の個数
	int num() {
		return _num;
	}
};

int diff(int d) {
	if (d < 0)d = 24 - (d % 24);
	d %= 24;
	cout << d << endl;
	return min(d, 24 - d);
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N;
	cin >> N;
	vi d(N);
	int ansmin = 25;
	rep(i, N) {
		cin >> d[i];
		if (d[i] > 12)d[i] = 24 - d[i];
		ansmin = min(ansmin, d[i]);
	}

	UnionFind uf(2 * N);

	rep(mind, 25) {
		rep(i, N - 1) {
			repr(j, i + 1, N) {
				if (abs(d[i] - d[j]) < mind) {
					uf.unite(i * 2, j * 2 + 1);
					uf.unite(1 * 2 + 1, j * 2);
				}
				if (abs(d[i] - (24 - d[j])) < mind) {
					uf.unite(i * 2, j * 2);
					uf.unite(i * 2 + 1, j * 2 + 1);
				}
			}
		}
		rep(i, N) {
			if (uf.same(i * 2, i * 2 + 1)) {
				cout << mind - 1 << endl;
				return 0;
			}
		}
	}

	cout << ansmin << endl;

	return 0;
}