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;
typedef long long ll;
typedef double db;
typedef pair<ll,ll> P;
#define pb push_back
#define ft first
#define sd second
#define mp make_pair
#define fr(i,n) for(int i=0;i<n;i++)
#define Fr(i,n) for(int i=0;i++<n;)
#define ifr(i,n) for(int i=n-1;i>=0;i--)
#define iFr(i,n) for(int i=n;i>0;i--)
ll n,d,e[13]={},c;

int f(ll c){
  ll ans=24,cnt=1;
  fr(i,23){
    if((c>>i)&1){
      ans=min(ans,cnt);
      cnt=1;
    }
    else cnt++;
  }
  return min(ans,cnt);
}

int rec(int i,ll c){
  //cout<<i<<" "<<(bitset<23>(c))<<endl;
  if(i==12){
    if(e[12]) return f(c+(1<<11));
    else return f(c);
  }
  if(e[i]==1) return max(rec(i+1,c+(1<<(i-1))),rec(i+1,c+(1<<(23-i))));
  if(e[i]==2) return rec(i+1,c+(1<<(i-1))+(1<<(23-i)));
  return rec(i+1,c);
}

int main(){
  cin>>n;
  fr(i,n){
    cin>>d;
    if(d==0){cout<<0<<endl;return 0;}
    if(++e[d]>2){cout<<0<<endl;return 0;}
    if(d==12&&e[d]>1){cout<<0<<endl;return 0;}
  }
  cout<<rec(1,0)<<endl;
}