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;

int main(){
    size_t N;
    cin >> N;

    vector<size_t> k(13), l(25);
    k[0] = 1;
    for(size_t i = 0, j; i < N; ++i){
        cin >> j;
        if(l[24 - j])return 0 & puts("0");
        l[j] = l[24 - j] = k[j];
        k[j] = 1;
    }

    vector<size_t> D, C;
    for(size_t i = 0; i < 13; ++i)if(k[i] ^ l[i])D.push_back(i);
    for(size_t i = 0; i < 25; ++i)if(l[i])C.push_back(i % 24);
    auto d = D.size(), c = C.size();

    vector<size_t> a(d + c);
    size_t ans = 0, K = 30;
    for(size_t m = 1; m < c; ++m)for(size_t n = 0; n < m; ++n)K = min(K, min(24 + min(C[m], C[n]) - max(C[m], C[n]), max(C[m], C[n]) - min(C[m], C[n])));
    for(size_t i = 0, j; i < 1 << d; ans = max(j, ans)){
        j = K;
        if(++i && d)(D[__builtin_ctzl(i)] *= -1) += 24;
        for(size_t m = 0; m < d; ++m)for(size_t n = 0; n < c; ++n)j = min(j, min(24 + min(D[m], C[n]) - max(D[m], C[n]), max(D[m], C[n]) - min(D[m], C[n])));
        for(size_t m = 1; m < d; ++m)for(size_t n = 0; n < m; ++n)j = min(j, min(24 + min(D[m], D[n]) - max(D[m], D[n]), max(D[m], D[n]) - min(D[m], D[n])));
    }

    cout << ans << endl;

    return 0;
}