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>

#define rep(i, a, b)	for(int i = a; i <= b; ++i)

using namespace std;

const int N = 55;

int n, D[N];
int F[N];
int cnt[N];
int mask;
int shit;

int calc(){
	int ans = 24;
	vector <int> showkat;
	for(int i = 0; i < 24; ++i) if(mask & (1 << i)) showkat.push_back(i);
	for(int i = 1; i < showkat.size(); ++i) ans = min(ans, showkat[i] - showkat[i - 1]);
	if(showkat.size() == 1) return 0;
	ans = min(ans, 24 - showkat.back() + showkat[0]);
	return ans;
}

void subha(int ans){
	cout << ans << endl;
	exit(0);
}

int main(){
	cin >> n;
	n += 1;
	cnt[0] += 1;
	for(int i = 2; i <= n; ++i){
		cin >> D[i];
		cnt[D[i]] += 1;
	}
	vector <int> F;
	for(int i = 0; i <= 12; ++i){
		if(cnt[i] > 2) subha(0);
		if(cnt[i] == 2 && i == 0) subha(0);
		if(cnt[i] == 2 || i == 0){
			mask |= 1 << i;
			mask |= 1 << (24 - i);
		}
		else if(cnt[i] == 1) F.push_back(i);
	}
	int gap = 0;
	n = F.size();
	for(int naz = 0; naz < (1 << n); ++naz){
		for(int i = 0; i < n; ++i){
			if(naz & (1 << i)) mask ^= 1 << F[i];
			else mask ^= 1 << (24 - F[i]);
		}
		gap = max(gap, calc());
		for(int i = 0; i < n; ++i){
			if(naz & (1 << i)) mask ^= 1 << F[i];
			else mask ^= 1 << (24 - F[i]);
		}
	}
	cout << gap << endl;
	return 0;
}
