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 <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <limits>
#include <map>
#include <regex>
#define REP(i,n) for(long (i)=0;(i)<(n);(i)++)
#define REPI(i,a,b) for(long (i)=(a);(i)<(b);(i)++)
#define INF numeric_limits<double>::infinity()
constexpr long MOD = 1e9 + 7;
using namespace std;
using P = pair<long, long>;
using VI = vector<long>;

int main() {
    int n;
    cin >> n;
    VI v(n + 1, 0);
    REP(i,n) {
        cin >> v[i + 1];
    }
    
    sort(v.begin(), v.end());
    REP(i,v.size()) {
        if (i % 2 == 0) {
            v[i] = 24 - v[i];
        }
    }
    
    long ans = 24;
    REP(i,v.size()) {
        REPI(j,i+1,v.size()) {
            long d = abs(v[j] - v[i]);
            ans = min(ans, min(d, 24 - d));
        }
    }
    cout << ans << endl;
}
