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 rep(i, n) for(int i = 0; i < (n); i++)
#define all(a) a.begin(), a.end()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
using namespace std;
typedef long long ll;

int main(){

    int N;
    cin >> N;

    vector<int> d(N);
    rep(i, N)cin >> d[i];
    d.push_back(0);
    sort(all(d));

    vector<int> cnt(24, 0);
    int now = 1;
    rep(i, N + 1){
        if(d[i] == 0 or d[i] == 12)cnt[d[i]]++;
        else {
            if(now == 1)cnt[d[i]]++;
            else cnt[24 - d[i]]++;
            now = 1 - now;
        }
    }

    int ans = 24;
    vector<int> tmp;
    rep(i, 24)rep(j, cnt[i])tmp.push_back(i);
    rep(i, N)chmin(ans, tmp[i + 1] - tmp[i]);
    cout << ans << endl;
    return 0;

}