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 d[51]={0}, t[51]={0};

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

    sort(d, d+n);

    for(int i=0;i<n;++i){
        t[i] = abs(24*(i%2) - d[i]);
    }

    int mini = d[0];
    for(int i=1;i<n;++i){
        for(int j=0;j<i;++j)
            mini = min(mini, min(abs(t[i]-t[j]), 24-abs(t[i]-t[j])));
    }

    cout << mini << endl;

    return 0;
}
