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 <iomanip>
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <utility>
#include <sys/time.h>
#include <cmath>
#include <tuple>
#include <queue>

using namespace std;

void solve(int n, vector<int> d) {
    sort(d.begin(), d.end());
    // int l = 0;
    //int r = 0;
    //int ret = 24;
    for (int i = 0; i < d.size(); i+=2) {
        d[i] = 24 - d[i];
    }
    sort(d.begin(), d.end());
    int ret = 24;
    int a = 0;
    for (int i = 0; i < d.size(); i++) {
        ret = min(ret, d[i] - a);
        a = d[i];
    }
    ret = min(ret, 24 - d.back());
    std::cout << ret << std::endl;
}

int main() {
    struct timeval start,end;
    long long span;
    int n;

    gettimeofday(&start,NULL);

    std::cin >> n;
    vector<int> d(n);
    for (int i = 0; i < n; i++) {
        std::cin >> d[i];
    }
    solve(n, d);

    gettimeofday(&end,NULL);
    span = (end.tv_sec -start.tv_sec)*1000000LL + (end.tv_usec - start.tv_usec);
    std::cerr << "--Total Time: " << span/1000 << "ms" << endl;
    return 0;
}
