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 n,cnt[13],ans;

bool f[24];

int main() {
    scanf("%d",&n);
    for (int i=1;i<=n;++i) {
        int t;
        scanf("%d",&t);
        cnt[t]++;
    }
    cnt[0]++;
    if (cnt[0]>=2 || cnt[12]>=2) return puts("0"),0;
    for (int i=1;i<=11;++i) if (cnt[i]>2) return puts("0"),0;
    for (int i=0;i<(1<<11);++i) {
        memset(f,0,sizeof f);
        f[0]=true;
        if (cnt[12]) f[12]=true;
        for (int j=1;j<=12;++j) {
            if (cnt[j]==1) {
                if (i&(1<<(j-1))) f[24-j]=true;
                else f[j]=true;
            } else if (cnt[j]==2) f[j]=f[24-j]=true;
        }
        vector<int> t;
        for (int j=0;j<23;++j) if (f[j]) t.push_back(j);
        for (int j=0,k=int(t.size());j<k;++j) t.push_back(t[j]+24);
        sort(t.begin(),t.end());
        int c=0x7fffffff;
        for (int j=0;j<int(t.size()-1);++j) c=min(c,t[j+1]-t[j]);
        ans=max(ans,c);
    }
    printf("%d\n",ans);
    return 0;
}