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>

#define fr first
#define sc second
#define OK puts("OK");
#define pb push_back
#define mk make_pair

using namespace std;

typedef long long ll;

const int inf = (int)1e9 + 7;
const int N = (int)1e5 + 7;

int n;
int a[N];
int u[N];

bool check (int x) {
        int cur = 1;
        vector <int> v;

        u[1] = 1;

        memset (u,0,sizeof(u));

        for (int i = 2; i <= n; i ++) {
                if (a[cur] + x >= a[i])
                        u[i] = 1,cur = i,v.pb(a[i]);
        }
        for (int i = 1; i <= n; i ++) {
                if (!u[i])
                        v.pb(24 - a[i]);
        }
        v.pb(v[0]);

        for (int i = 0; i < v.size() - 1; i ++) {
                if (min(abs(v[i] - v[i + 1]),24 - abs(v[i] - v[i + 1])) < x)
                        return false;
        }
        return true;

}
int main () {
        cin >> n;

        if (n == 1) {
                cout << 0;
                return 0;
        }

        for (int i = 1; i <= n; i ++)
                cin >> a[i];

        sort (a + 1,a + n + 1);

        for (int i = 12; i >= 0; i --) {
                if (check(i)) {
                        cout << i << endl;
                        return 0;
                }
        }
}