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;

typedef long long int ll;
typedef pair<ll, ll> pll;

#define FOR(i, n, m) for (ll(i) = (m); (i) < (n); ++(i))
#define REP(i, n) FOR(i, n, 0)
#define OF64 std::setprecision(10)

const ll MOD = 1000000007;
const ll INF = (ll)1e15;

int T[15];

int get(ll n)
{
    ll min = 12;
    REP(i, 24)
    {
        FOR(j, 24, i + 1)
        {
            if (((n >> i) & 1) == 0 || ((n >> j) & 1) == 0)
                continue;
            min = std::min(min, std::min(j - i, 24 + i - j));
        }
    }
    return min;
}

int solve(ll n = 0, int index = 0)
{
    if (index > 12)
        return get(n);

    if (T[index] == 0)
        return solve(n, index + 1);

    ll n0 = 1 << index;
    ll n1 = 1 << ((24 - index) % 24);
    if (T[index] == 2)
        return solve(n | n0 | n1, index + 1);
    return std::max(solve(n | n0, index + 1), solve(n | n1, index + 1));
}

int main()
{
    int N;
    cin >> N;
    T[0] = 1;
    REP(i, N)
    {
        int d;
        cin >> d;
        T[d]++;
    }

    bool isZero = false;
    REP(i, 13)
    {
        if (T[i] > 2)
            isZero = true;
    }
    if (T[0] > 1 || T[12] > 1)
        isZero = true;

    int ans = 0;
    if (!isZero)
    {
        ans = solve();
    }
    cout << ans << endl;
    return 0;
}