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;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(x) x.begin(),x.end()
#define dbg(x) cout<<#x<<":"<<x<<endl
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<ll,P> PP;

int ans;
int cnt[13];
int flag[25];

void dfs(int tn=1){
  if(tn==12){
    int rn=0;
    int mi=100;
    for(int i=1;i<25;i++){
	 if(flag[i]==0)rn++;
	 else {
	   mi=min(mi,rn);
	   rn=0;
	 }
    }
    ans=max(ans,mi+1);
    return;
  }

  if(cnt[tn]==1){
    flag[tn]=1;
    dfs(tn+1);
    flag[tn]=0;
    flag[24-tn]=1;
    dfs(tn+1);
    flag[24-tn]=0;
    return;
  }
  
  dfs(tn+1);
}

int main(){
  int n,a[55];
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>a[i];
  }

  cnt[0]=2;
  cnt[12]=1;
  for(int i=0;i<n;i++){
    cnt[a[i]]++;

    if(cnt[a[i]]==2){
	 flag[a[i]]=1;
	 flag[24-a[i]]=1;
    }
    
    if(cnt[a[i]]==3){
	 cout<<0<<endl;
	 return 0;
    }
    if(a[i]==12)flag[12]=1;
  }

  flag[0]=flag[24]=1;
  dfs();
  cout<<ans<<endl;
  return 0;
}
