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 <string>
#include <algorithm>
#include <utility>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define rep(i, n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
const ll mod = 1000000007;

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

    vector<int> d(n);
    rep(i, n) {
        cin >> d[i];
    }
    sort(d.begin(), d.end());

    vector<int> t(n+1);
    rep(i, n) {
        if(i%2 == 0) {
            t[i] = d[i];
        }
        else {
            t[i] = 24 - d[i];
        }
    }

    int ans = 30;
    rep(i, n) {
        for(int j = i+1; j <= n; j++) {
            int x = abs(t[i] - t[j]);
            int y = min(x, 24-x);
            ans = min(ans, y);
        }
    }

    cout << ans << endl;
    return 0;
}