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;
using ll = int64_t;
int cnt[12 + 1] = {1};

int dfs(int num, vector<int> v){
	if(num == 13){
		sort(v.begin(), v.end());
		int ret = 1e5;
		for(int i = 0; i < v.size() - 1; ++i) ret = min(ret, min((int)abs(v[i] - v[i + 1]), 24 - (int)abs(v[i] - v[i + 1])));
		ret = min(ret, min((int)abs(v[v.size() - 1] - v[0]), 24 - (int)abs(v[v.size() - 1] - v[0])));
		return ret;
	}
	if(!cnt[num]){
		return dfs(num + 1, v);
	}else if(cnt[num] == 2){
		v.push_back(num);
		v.push_back(24 - num);
		return dfs(num + 1, v);
	}else{
		vector<int> tmp = v;
		tmp.push_back(num);
		int ret = dfs(num + 1, tmp);
		v.push_back(24 - num);
		ret = max(ret, dfs(num + 1, v));
		return ret;
	}
}

int main(){
	int n; cin >> n;
	int d[n];
	for(int i = 0; i < n; ++i) cin >> d[i];
	for(int i = 0; i < n; ++i){
		++cnt[d[i]];
		if(cnt[d[i]] >= 3 || !d[i]){
			cout << 0 << endl;
			return 0;
		}
	}
	vector<int> v;
	v.push_back(0);
	cout << dfs(1, v) << endl;
	return 0;
}