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>
#include <algorithm>
using namespace std;
int main(){
	int N; 
	int D[15] = {}; 
	D[0] = 1; 
	cin >> N; 
	for(int i = 0; i < N; i++){
		int a; cin >> a; D[a] ++; 
	}
	int ans = 12; 
	for(int i = 1; i < 12; i++){
		if(D[i] >= 3) ans = 0; 
	}
	if(D[0] >= 2) ans = 0; 
	if(D[12] >= 2) ans = 0; 
	if(ans > 0){
		vector<int> exist = {0, 24};  
		for(int i = 1; i < 12; i++){
			if(D[i] == 2){
				exist.push_back(i); 
				exist.push_back(24 - i); 
			}
			else if(D[i] == 1){
				if(exist.at(exist.size()-1) < 12){
					exist.push_back(24 - i); 
				}
				else{
					exist.push_back(i); 
				}
			}
		}
		if(D[12] == 1) exist.push_back(12); 
		sort(exist.begin(), exist.end()); 
		for(int j = 0; j < exist.size() - 1; j++){
			ans = min(ans, exist[j+1] - exist[j]); 
		}
 	}
	cout << ans << endl; 
}