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 ll;
const ll INF = 1LL << 60;
const ll MOD = 1e9 + 7;
#define rep(i,n) for (int i = 0; i < (n); i++)
#define ALL(v) v.begin(), v.end()
int cnt[13];
int ans;
void solve(int t, int bit) {
	if (t == 13) {
		int mi = 1 << 30;
		for (int i = 0; i < 24; i++) {
			if ((bit&(1 << i)) == 0) continue;
			for (int j = i + 1; j < 24; j++) {
				if ((bit&(1 << j)) == 0) continue;
				int d = abs(i - j);
				mi = min(mi, min(d, 24 - d));
			}
		}
		ans = max(ans, mi);
		return;
	}
	if (cnt[t] == 0) {
		solve(t + 1, bit);
		return;
	}
	if (cnt[t] == 1) {
		solve(t + 1, bit | (1 << t));
		if (t > 0) solve(t + 1, bit | (1 << (24 - t)));
		return;
	}
	solve(t + 1, bit | (1 << t) | (1 << (24 - t)));
}
int main() {
	int N;
	cin >> N;
	cnt[0]++;
	rep(i, N) {
		int D;
		cin >> D;
		cnt[D]++;
	}
	if (cnt[0] > 1 || cnt[12] > 1) {
		cout << 0 << endl;
		return 0;
	}
	for (int i = 1; i < 12; i++) {
		if (cnt[i] > 2) {
			cout << 0 << endl;
			return 0;
		}
	}
	solve(0, 0);
	cout << ans << endl;
}