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>
#define For(i, a, b) for(int (i)=(a); (i)<(b); ++(i))
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
using namespace std;
typedef long long lint;

int main(){
    int n;
    scanf("%d", &n);
    int t[13];
    rep(i, 13) t[i] = 0;
    ++t[0];
    rep(i, n){
        int d = 0;
        scanf("%d", &d);
        ++t[d];
    }
    vector<int> a, b;
    rep(i, 13){
        if(t[i] > 2){
            printf("%d\n", 0);
            return 0;
        }
        if(t[i] == 2){
            a.push_back(i);
            a.push_back(-i);
        }
        if(t[i] == 1) b.push_back(i);
    }
    int ans = 0;
    int s = 30;
    rep(i, a.size())For(j, i+1, a.size()) s = min(s, min(abs(a[i]-a[j]), 24-abs(a[i]-a[j])));
    rep(i, 1<<b.size()){
        int tmp = min(30, s);
        rep(j, b.size())if((i>>j)&1) b[j] *= -1;
        rep(j, b.size()){
            For(k, j+1, b.size()) tmp = min(tmp, min(abs(b[k]-b[j]), 24-abs(b[k]-b[j])));
            rep(k, a.size()) tmp = min(tmp, min(abs(a[k]-b[j]), 24-abs(a[k]-b[j])));
        }
        ans = max(ans, tmp);
        rep(j, b.size()) b[j] = abs(b[j]);
    }
    printf("%d\n", ans);
}