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<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

int main(){
    int N, D[51] = {}, pos[51] = {};
    cin >> N;
    for(int i = 1; i <= N; i++){
        cin >> D[i];
    }
    sort(D, D+(N+1));
    bool right = true;
    for(int i = 0; i <= N; i++){
        if(right){
            pos[i] = D[i];
        }else{
            pos[i] = (24-D[i])%24;
        }
        right = !right;
    }
    sort(pos, pos+(N+1));
    for(int i = 0; i <= N; i++){
        //printf("%d: %d\n", i, pos[i]);
    }
    int ans = 10000;
    for(int i = 0; i < N; i++){
        ans = min(ans, pos[i+1] - pos[i]);
    }
    ans = min(ans, 24 - pos[N]);
    cout << ans << endl;
}


        
    