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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
#include <deque>
#include <iomanip>
#include <limits>
#include <fstream>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
bool ex[24];
int cnt[13];
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;cin>>N;
  ex[0]=true;
  FOR(i,0,N){
    int d;cin>>d;
    cnt[d]++;
  }
  if(cnt[0]||cnt[0]>=2||cnt[12]>=2){
    cout<<0<<endl;
    return 0;
  }
  if(cnt[12])ex[12]=true;
  vector<int> v;
  FOR(i,1,12){
    if(cnt[i]==1){
      v.push_back(i);
    }
    if(cnt[i]==2){
      ex[i]=ex[24-i]=true;
    }
    if(cnt[i]>=3){
      cout<<0<<endl;
      return 0;
    }
  }
  int ans=0;
  int n=v.size();
  FOR(i,0,1<<n){
    bool ex2[24];
    FOR(j,0,24) ex2[j]=ex[j];
    FOR(j,0,n){
      if(i&(1<<j)){
        ex2[v[j]]=true; 
      }else{
        ex2[24-v[j]]=true;
      }
    }
    int mn=1e9;
    FOR(i,0,24){
      FOR(j,i+1,24){
        if(ex2[i] && ex2[j]){
          mn=min(mn,abs(j-i));
          mn=min(mn,24-abs(j-i));
        }
      }
    }
    if(mn!=1e9)ans=max(ans,mn);
  }
  cout<<ans<<endl;
}