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 <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <string>
using namespace std;
#define LL long long
#define lc c[x][0]
#define rc c[x][1]
#define pb push_back
int cnt[13],D[52],d[52],n,ans=0;
int cal(int a,int b){
    if (a<=0&&b<=0) a*=-1,b*=-1;
    if (a>b) swap(a,b);
    if (a<0) return min(b-a,24-(b-a));
     else return (b-a);
}
void dfs(int i){
    if (i==n+1){
        int j,k,mx=24;
        for (j=0;j<=n;j++)
         for (k=j+1;k<=n;k++)
          mx=min(mx,cal(d[j],d[k]));
        ans=max(ans,mx);
        return;
    }
    d[i]=D[i];dfs(i+1);
    d[i]=-D[i];dfs(i+1);
}
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int i;
    cin>>n;cnt[0]++;
    for (i=1;i<=n;i++){
        scanf("%d",&D[i]);
        cnt[D[i]]++;
    }
    if (cnt[0]>1||cnt[12]>1) {puts("0");return 0;}
    for (i=1;i<=11;i++) if (cnt[i]>2) {puts("0");return 0;}
    for (i=1;i<=11;i++) if (cnt[i]==2) if (cnt[i-1]>0||cnt[i+1]>0) {puts("1");return 0;}

    d[0]=0;dfs(1);
    cout<<ans<<endl;
    return 0;
}
