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 INF (sizeof(int) == 4 ? (int)1e9:(int)1e18)
template<class T, class S> void cmin(T &a, const S &b) { if (a > b)a = b; }
template<class T, class S> void cmax(T &a, const S &b) { if (a < b)a = b; }
signed main(){
    int N; cin >> N;
    vector<int>v(N);
    for(auto & e : v) cin >> e;
    array<short,13> cnt = {0};
    cnt[0]++;
    for(auto  e : v)cnt[e]++;
    for(auto  e : cnt) if(e >= 3) return puts("0") * 0;
    if(cnt[0] >= 2 or cnt[12] >= 2) return puts("0")* 0;
    int ans = -INF;
    
    for(int bit = 0; bit < (1LL << 12); ++bit){
        vector<int>pos(25,0LL);
        pos[0] = cnt[0];
        pos[12] = cnt[12];
        int tans = INF;
        for(int i = 1; i <= 12; ++i){
            if(cnt[i] == 0) continue;
            if(cnt[i] == 2) pos[i] = pos[24 - i] = 1;
            if(cnt[i] == 1){
                ((bit & (1LL << (i - 1))) >= 1 ? pos[i] : pos[24 - i]) = 1;
            }
        }
        for(int i = 0; i < pos.size(); ++i){
            for(int j = 0; j < pos.size(); ++j){
                if(i != j and pos[i] == 1 and pos[j] == 1)
                    cmin(tans, min(abs(i-j),24 - abs(i-j)));
            }
        }
        if(tans == INF)continue;
        cmax(ans, tans);
    }
    cout << ans << endl;
}
