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 pair<ll,ll> P;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;cin>>n;
    vector<int> d(n+1,0),pos(n+1,0);
    for(int i=1;i<=n;i++) cin>>d[i];
    sort(d.begin(),d.end());
    bool f=1;
    for(int i=1;i<=n;i++){
        if(f) pos[i]=-d[i];
        else pos[i]=d[i];
        f = !f;
        //cout<<pos[i]<<endl;
    }
    int res=100;
    for(int i=0;i<=n;i++){
        for(int j=i+1;j<=n;j++){
            int dif=abs(max(pos[i],pos[j])-min(pos[i],pos[j]));
            res = min(res,min(dif,24-dif));
        }
    }
    cout<<res<<endl;
}
