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 <bits/stdc++.h>
using namespace std;

int A[23], B[23];
set<int> s;
set<int>::iterator it;

int main() {
    int N, x, t = 0, r, i, j;
    scanf("%d", &N);
    if (N > 23) {printf("0\n"); return 0;}
    if (N == 23) {
        for (i = 0; i < N; ++i) scanf("%d", &A[i]);
        sort(A, A + N);
        for (i = 0; i < N; ++i) if (A[i] != i / 2 + 1) {printf("0\n"); return 0;}
        printf("1\n");
        return 0;
    }
    for (i = 0; i < N; ++i) scanf("%d", &A[i]);
    for (i = 0; i < (1 << N); ++i) {
        s.insert(0);
        s.insert(24);
        r = 24;
        for (j = 0; j < N; ++j) {
            x = (i & (1 << j)) ? 24 - A[j] : A[j];
            it = s.upper_bound(x);
            if (it == s.end()) {
                printf("0\n");
                return 0;
            }
            r = min(r, abs(*it - x));
            r = min(r, abs(x - *(--it)));
            s.insert(x);
        }
        s.clear();
        t = max(t, r);
    }
    printf("%d\n", t);
    return 0;
}
