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>

using namespace std;
typedef long long int ll;
typedef pair <ll,ll> P;
#define REP(i, n) for(ll i = 0LL; i < (ll)(n); i++)
#define REPR(i, n) for(ll i = n-1; i >= 0LL; i--)
#define FOR(i, n, m) for(ll i = n; i < (ll)(m); i++)
#define PRINT(x) cout << x << endl
#define ALL(v) v.begin(), v.end()
#define INF (ll)1e18
#define INFD 1e18
#define PI 3.14159265358979
#define MOD 1000000007
#define vl vector<ll>

vector<ll> dx = {-1, 1, 0, 0};
vector<ll> dy = {0, 0, 1, -1};

ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

int main()
{
	ll N;
	cin >> N;
	vector<ll> D(N);
	REP(i, N) cin >> D[i];
	vector<ll> v;
	v.push_back(0);
	ll MAX = INF;
	REP(i, N) {
		ll MIN1 = INF;
		v.push_back(D[i]);
		REP(j, v.size()) {
			FOR(k, j+1, v.size()) {
				MIN1 = min(MIN1, abs(v[j] - v[k]));
			}
		}
		v.pop_back();
		ll MIN2 = INF;
		v.push_back((24 - D[i]) % 24);
		REP(j, v.size()) {
			FOR(k, j+1, v.size()) {
				MIN2 = min(MIN2, abs(v[j] - v[k]));
			}
		}
		v.pop_back();
		if (MIN1 > MIN2) {
			v.push_back(D[i]);
		} else {
			v.push_back((24 - D[i])%24);
		}
	}
	REP(i, v.size()) {
		FOR(j, i+1, v.size()) {
			MAX = min(MAX, abs(v[i]-v[j]));
		}
	}
	PRINT(MAX);
	return 0;
}