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;

int h[13];

int f[15][30][30];

int main(){
    int n;
    cin>>n;
    h[0]=2;
    while (n--){
        int val;
        cin>>val;
        h[val]++;
    }
    if ((*max_element(h,h+13))>2){
        cout<<0;
        return 0;
    }
    f[0][12][12]=30;
    for(int cur=0;cur<12;cur++) {
        if (h[cur+1]==1){
            for(int L=12-cur;L<=12;L++)
                for(int R=12;R<=cur+12;R++) if (f[cur][L][R]){
//                    cout<<cur<<" "<<L<<" "<<R<<" "<<f[cur][L][R]<<'\n';
                    f[cur+1][L][12+cur+1]=max(f[cur+1][L][12+cur+1],min(f[cur][L][R],12+cur+1-R));
                    f[cur+1][12-cur-1][R]=max(f[cur+1][12-cur-1][R],min(f[cur][L][R],L-(12-cur-1)));
                }
            continue;
        }
        if (h[cur+1]==2){
            for(int L=12-cur;L<=12;L++)
                for(int R=12;R<=cur+12;R++) if (f[cur][L][R])
                    f[cur+1][12-cur-1][12+cur+1]=max(f[cur+1][12-cur-1][12+cur+1],min(min(L-(12-cur-1),(12+cur+1)-R),f[cur][L][R]));
            continue;
        }
//        cout<<"tick\n";
        for(int L=12-cur;L<=12;L++)
            for(int R=12;R<=cur+12;R++) f[cur+1][L][R]=f[cur][L][R];
    }
    int ans=0;
    for(int L=0;L<30;L++)
        for(int R=0;R<30;R++) ans=max(ans,min(f[12][L][R],24-(R-L)));
    cout<<ans;
}
