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: Accepted)

#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <cstring>
#include <numeric>
#include <algorithm>
#include <functional>
#include <array>
#include <map>
#include <queue>
#include <limits.h>
#include <set>
#include <stack>
#include <random>
#include <complex>
#include <unordered_map>
#define rep(i,s,n) for(int i = (s); (n) > i; i++)
#define REP(i,n) rep(i,0,n)
#define RANGE(x,a,b) ((a) <= (x) && (x) < (b))
#define DUPLE(a,b,c,d) (RANGE(a,c,d) || RANGE(b,c,d) || RANGE(c,a,b) || RANGE(d,a,b))
#define INCLU(a,b,c,d) (RANGE(a,c,d) && (b,c,d))
#define PW(x) ((x)*(x))
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define MODU 1000000007LL
#define bitcheck(a,b)   ((a >> b) & 1)
#define bitset(a,b)      ( a |= (1 << b))
#define bitunset(a,b)    (a &= ~(1 << b))
#define MP(a,b) make_pair((a),(b))
#define Manh(a,b) (abs((a).first-(b).first) + abs((a).second - ((b).second))
#define pritnf printf
#define scnaf scanf
#define itn int
#define PI 3.141592653589


#define izryt bool
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
template<typename A, size_t N, typename T>
void Fill(A(&array)[N], const T &val) {
	std::fill((T*)array, (T*)(array + N), val);
}
pll Dir[8] = { //移動
	{ 0 ,1 },{ -1 ,0 },{ 1 ,0 },{ 0 ,-1 },
	{ 1 ,1 },{ 1 ,-1 },{ -1 ,1 },{ -1 ,-1 }
};

//[a, b)
#define Getsum(ar, a,b) (ar[b] - ar[a])
#define INF 10000000000000000LL

struct Edge {
	int from, to;
	ll w;
};
typedef vector<vector<Edge>> Graph;

ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a%b);
}
ll lcm(ll m, ll n)
{
	return m * n / gcd(m, n);
}

ll gettugibaisu(ll from, ll k) {

	ll ret = ((from + k - 1) / k) * k;

	return ret;
}

ll getmaebaisu(ll to, ll k) {

	ll ret = (to / k) * k;

	return ret;
}


signed main(void) {

	int n;
	cin >> n;
	vector<int> d(n);
	vector<pii> tl;
	REP(i, n) {
		cin >> d[i];
		if (d[i] == 0) {
			cout << 0 << endl;
			return 0;
		}
	}

	REP(i, n) {
		tl.push_back({ d[i], i + 1 });
			tl.push_back({24 - d[i], -(i + 1)});
		}
	

	sort(ALL(tl));

	if (n >= 24) {
		cout << 0 << endl;
		return 0;
	}
	int Max = 0;
	int tori = pow(2, n);
	REP(i, tori) {
		int Min = 300;
		int bef = 0;
		int ar[24] = {};
		int buf = i;
		REP(j, n) {
			ar[j] = buf & 1;
			buf >>= 1;
		}

		for (auto itr : tl) {
			if (itr.second > 0) {
				if (ar[itr.second - 1] == 0) {
					Min = min(Min, itr.first - bef);
					bef = itr.first;
				}
			}
			else {
				if (ar[-itr.second - 1] == 1) {
					Min = min(Min, itr.first - bef);
					bef = itr.first;
				}
			}
		}
		Min = min(Min, 24 - bef);
		Max = max(Max, Min);
	}
	
	cout << Max << endl;

	return 0;
}