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 REPI(i, n) for(int i = 0; i < n; i++)
#define REPJ(j, n) for(int j = 0; j < n; j++)

using namespace std;

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

    int n; cin >> n;
    vector<int> vc;
    vc.push_back(0);
    for(int i = 0; i < n; i++){
        int x; cin >> x;
        vc.push_back(x);
    }
    sort(vc.begin(), vc.end());

    for(int i = 0; i <= n; i+=2){
        vc[i] = (24 - vc[i]) % 24;
    }

    sort(vc.begin(), vc.end());

    int minimum = 100;

    for(int i = 1; i <= n; i++){
        minimum = min(vc[i] - vc[i - 1], minimum);
    }
    cout << minimum;
}
