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 rep(i,n) for(int i=0;i<int(n);++i)
#define debug(x) cout<<#x<<": "<<x<<endl

int n;
map<int, int> mp;
int res = 1;
bool det[24]={};

void rec(int k){
    if(k==12){
        int ans = 12;
        rep(i,24)rep(j,i){
            if(det[i] && det[j]){
                ans = min(ans,min(24-i+j,i-j));
            }
        }
        res = max(res,ans);
        return;
    }
    if(mp[k]==1){
        det[k] = true;
        rec(k+1);
        det[k] = false;
        det[24-k] = true;
        rec(k+1);
        det[24-k] = false;
    }else{
        rec(k+1);
    }
}

int solve(){
    cin>>n;
    mp[0]++;
    rep(i,n){
        int d;
        cin>>d;
        mp[d]++;
        if(mp[d] >= 3) return 0;
    }
    if(mp[0] >= 2 || mp[12] >= 2) return 0;
    
    det[0] = true;
    if(mp[12]==1) det[12] = true;
    rep(i,12) if(mp[i]==2) det[i] = det[24-i] = true;
    rec(1);
    return res;
}

int main(void){
    cout<<solve()<<endl;
    return 0;
}
