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.

Examples

Input

3
7 12 8


Output

4


Input

2
11 11


Output

2


Input

1
0


Output

0
#include <iostream>
#include <vector>
#include <algorithm>
#include <tuple>

int solve(int N, std::vector<int> D)
{
    N++;
    D.emplace_back(0);
    
    if (N > 24) {
        return 0;
    }
    
    std::vector<int> cnt(13);
    for (int i = 0; i < N; i++) {
        cnt[D[i]]++;
        if (cnt[D[i]] == 3) {
            return 0;
        }
    }

    if (N > 12) {
        return 1;
    }
    
    int ret = 0;
    for (int i = 0; i < (1 << N); i++) {
        std::vector<int> d(N, 0);
        for (int j = 0; j < N; j++) {
            if (i >> j & 1) {
                d[j] = D[j] % 24;
            } else {
                d[j] = (24 - D[j]) % 24;
            }
        }
        
        int score = 10000000;
        for (int j = 0; j < N; j++) {
            for (int k = j + 1; k < N; k++) {
                int min, max;
                std::tie(min, max) = std::minmax(d[j], d[k]);
                
                int diff = std::min((max - min) % 24, (24 - (max - min)) % 24);                
                score = std::min(score, diff);
            }
        }

        
        ret = std::max(ret, score);
    }
    
    return ret;
}

int main()
{
    int N;
    std::cin >> N;
    
    std::vector<int> D(N);
    for (int i = 0; i < N; i++) {
        std::cin >> D[i];
    }
    
    std::cout << solve(N, D) << std::endl;

    return 0;
}
