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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define fi first
#define se second
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const ll INF = 1ll<<30;
const ll longINF = 1ll<<60;
const ll MOD = 1000000007;
const double EPS = 1e-9;
const bool debug = 0;
//---------------------------------//

int N;
int D[50];
bool dp[2][1<<24];

int main() {
	cin >> N;
	REP(i, N) scanf("%d", D + i);
	
	dp[0][0] = true;
	REP(i, N) {
		memset(dp[i&1^1], 0, sizeof(dp[0]));
		REP(s, 1<<24) if (dp[i&1][s]) {
			dp[i&1^1][s | 1<<D[i]] = true;
			dp[i&1^1][s | 1<<(24-D[i])] = true;
		}
	}
	
	int ans = 0;
	REP(s, 1<<24) if (dp[N&1][s]) {
		int cur = 100;
		REP(i, 24) if (s >> i & 1) {
			int cnt = 1;
			while (i + cnt < 48 && !(s >> ((i + cnt) % 24) & 1)) cnt++;
			
			if (i + cnt >= 24) break;
			i = i + cnt - 1;
			
			chmin(cur, cnt);
		}
		
		if (cur < 24) chmax(ans, cur);
	}
	cout << ans << endl;
	return 0;
}