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>
using namespace std;
int n, cnt[13];
bool H[25];

int count() {
    int i = 0, j = -1, ans = 24;
    while (i < 25) {
        if (H[i]) {
            if (j >= 0) ans = min(ans, i-j);
            j = i;
        }
        i++;
    }
    return ans;
}

int solve(int h) {
    if (h == 12) {
        if (cnt[h] > 0) H[h] = true;
        int a = count();
        H[h] = false;
        return a;
    } else {
        if (cnt[h] == 0) return solve(h+1);
        else if (cnt[h] == 1) {
            H[h] = true;
            int a1 = solve(h+1);
            H[h] = false;
            H[24-h] = true;
            int a2 = solve(h+1);
            H[24-h]= false;
            return max(a1, a2);
        } else if (cnt[h] == 2) {
            H[h] = H[24-h] = true;
            int a = solve(h+1);
            H[h] = H[24-h] = false;
            return a;
        } else return 0;
    }
}

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        int h; cin >> h;
        cnt[h]++;
    }
    H[0] = H[24] = true;
    if (cnt[0] > 0 || cnt[12] > 1) { cout << "0\n"; }
    else cout << solve(1) << endl;
}
