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<set>
#include <bitset>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include <cstdio>
#include<algorithm>
#include <sstream>
#include<string>
#include<string.h>
#include <cmath>
#include <iomanip>
#include <string>
#include<list>
#include <limits>
#include <numeric>
#include <type_traits>
#define int long long
#define ll long long
#define mod  1000000007
#define inf 1e17
#define rep(i,j,n) for(int i=j;i<n;i++)
#define P pair<int,int>
double pi = 3.141592653589793;
using namespace std;
//ここから始めよう
signed main(){
    int n;cin>>n;
    int sum[13]={};sum[0]=1;
    vector<int> num;
    rep(i,0,n){
        int d;cin>>d;
        if(d==0){
            cout<<0<<endl;
            return 0;
        }
        num.push_back(d);
        sum[d]++;
    }
    sort(num.begin(),num.end());
    rep(i,1,12){
        if(sum[i]>=3){
            cout<<0<<endl;
            return 0;
        }
    }
    if(sum[12]>=2){
        cout<<0<<endl;
        return 0;
    }
    int place[24]={};place[0]=1;
    int c=1;
    rep(i,0,n){
        if(c==1){
            place[num[i]]=1;
        }else {
            place[24-num[i]]=1;
        }c*=-1;
    }
    int count=0;int ans=114514;
    rep(i,1,24){
        if(place[i]==0)count++;
        else {
            ans=min(ans,count);
            count=0;
        }
    }cout<<ans+1<<endl;
    return 0;
}