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;
string s;
int main(void) {
    std::ios::sync_with_stdio(0);
    cin.tie(), cout.tie(0);
    int a[55] = {0}, n;
    while(cin >> n) {
        for(int i = 1; i <= n; ++i)
            cin >> a[i];
        sort(a + 1, a + n + 1);
        for(int i = n; i >= 1; --i) {
            if(!(i & 1)) a[i] = 24 - a[i];
        }
        sort(a + 1, a + n + 1);
        int mn = 24;
        for(int i = 1; i <= n; ++i)
            mn = min(mn, a[i] - a[i - 1]);
        cout << mn << endl;
    }
}