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;

long long N;
long long pos[100];
bool there[24];
long long ans = 0;


long long dfs(long long idx){
	if(idx == 24){
		vector<long long>vec;
		vec.push_back(0);
		for(long long i = 0; i < 24; i++){
			if(there[i]){
				vec.push_back(i);
			}
		}
		vec.push_back(24);

		long long tmp = INT_MAX;
		for(long long i = 0; i < vec.size() -1 ; i++){
			tmp = min(tmp, vec[i+1] - vec[i]);
		}
		ans = max(tmp, ans);
		return 0;
	}

	if(pos[idx] == 2){
		there[idx] = true;
		there[24-idx] = true;
		dfs(idx + 1);
		there[idx] = false;
		there[24-idx] = false;
		return 0;
	}else if(pos[idx] == 1){
		there[idx] = true;
		dfs(idx + 1);
		there[idx] = false;
		there[24-idx] = true;
		dfs(idx + 1);
		there[24-idx] = false;
		return 0;
	}else if(pos[idx] == 0){
		dfs(idx + 1);
	}else{
		exit(1);
	}
}


int main(){
	cin >> N;
	for(long long i = 0; i < N; i++){
		long long tmp;
		cin >> tmp;
		pos[tmp]++;
		ans = min(ans, tmp);
	}
	if(pos[0] > 0){
			cout << 0 << endl;
			return 0;
	}

	for(long long i = 0; i <= 12; i++){
		if(pos[i] >= 3){
			cout << 0 << endl;
			return 0;
		}
	}

	dfs(0);
	cout << ans << endl;

}