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;

#define all(x) begin(x), end(x)

using i64 = int_fast64_t;
using f64 = double;
using p64 = pair<i64, i64>;

constexpr i64 INF = INT_FAST64_MAX / 2;
constexpr f64 EPS = DBL_EPSILON;

void solve() {
    i64 n;
    cin >> n;
    vector<i64> d(n);
    for (auto& e : d) cin >> e;
    sort(all(d));

    for (i64 i = 0; i < n; i++) {
        if (i % 2) d[i] = -d[i];
    }
    sort(all(d));

    i64 ans = d.back() - d.front();
    ans = min(ans, 24 - ans);
    for (i64 i = 0; i + 1 < n; i++) {
        ans = min(ans, d[i + 1] - d[i]);
    }
    cout << ans << endl;
}

int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    solve();
}