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>
#define p(s) cout<<(s)<<endl
#define REP(i,n,N) for(int i=n;i<N;i++)
#define RREP(i,n,N) for(int i=N-1;i>=n;i--)
#define CK(n,a,b) ((a)<=(n)&&(n)<(b))
#define F first
#define S second
typedef long long ll;
using namespace std;
const int inf = 1e9;

int D[55],N;
string s;
int s1,s2;
int main() {
    cin>>N;
    REP(i,0,N){
        cin>>D[i];
    }
    D[N]=0;
    sort(D,D+N+1);

    REP(i,0,N+1){
        if(i%2){
            D[i]=(24-D[i])%24;
        }
    }
    
    int mn=23;
    REP(i,0,N+1) REP(j,i+1,N+1){
        int d = abs(D[i]-D[j]);
        mn=min(mn,min(d,24-d));
    }
    p(mn);
	return 0;
}
