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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <sys/time.h>

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <utility>
#include <cctype>

using namespace std;

using ll = long long;
using P = pair<double, double>;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define mkp(a, b) make_pair(a, b)
ll mod = 1000000007LL;

const ll inf = 1e9;

int diff(int i, int j) {
	int ans = max(abs(i - j), abs((24 - i) - j));
	ans = max(ans, abs(i - (24 - j)));
	ans = max(ans, abs((24 - i) - (24 - j)));

	return min(ans, 24 - ans);
}

int main() {
	int n;
	cin >> n;

	vector<int> DS(13,0);
	rep(i,n) {
		int d;
		cin >> d;
		DS[d]++;
	}
	DS[0]++;

	int min_v = 24;

	for(int i = 0; i <= 12; i++) {
		for (int j = i; j <= 12; j++) {
			if ((i == j && DS[i] >= 2) || (i != j && DS[i] > 0 && DS[j] > 0)) {
				min_v = min(min_v, diff(i,j));
			}
		}
	}
	cout << min_v << endl;




	return 0;
}