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;
#define rep(i,n) for (int (i)=(0);(i)<(int)(n);++(i))
using ll = long long;
using P = pair<int, int>;
using namespace std;

template<class T> void vin(vector<T>& v, int n) {
    v.resize(n);
    for (int i = 0; i < n; ++i) {
        cin >> v[i];
    }
}

int main() {
    int n;
    cin >> n;
    vector<int> d, v;
    v.push_back(0);
    rep(i, n) {
        int tmp;
        cin >> tmp;
        d.push_back(tmp);
    }
    sort(d.begin(), d.end());
    rep(i, n) {
        v.push_back(d[i] * (i%2 ? -1 : 1));
    }
    sort(v.begin(), v.end());
    int ans = 1e8;
    rep(i, n) {
        ans = min(ans, v[i+1] - v[i]);
    }
    ans = min(ans, abs(v[0] + 24 - v[n]));
    cout << ans << endl;
    return 0;
}
