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<iostream>
#include<algorithm>
#include<cassert>
#include<vector>
#include<fstream>
#include<string>
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<queue>
#include<map>

using namespace std;
#define REP(i,m,n) for(long long i=m;i<n;++i)
#define INF 10000000000000000
#define MOD 1000000007
#define ll long long
#define mp make_pair
#define all(v) v.begin(),v.end()

vector<int>v(13, 0);
vector<int>ans, hoge;

int main() {
	int n;
	cin >> n;
	v[0]++;
	REP(i, 0, n) {
		int tmp;
		cin >> tmp;
		v[tmp]++;
	}
	REP(i, 1, 12) {
		if (v[i] > 2) {
			cout << 0 << endl; return 0;
		}
	}
	if (v[0] >= 2) {
		cout << 0 << endl; return 0;
	}
	if (v[12] >= 2) {
		cout << 0 << endl; return 0;
	}
	REP(i, 1, 12) {
		if (v[i] == 2) {
			if (v[i - 1] == 1 || v[i + 1] == 1) {
				cout << 1 << endl; return 0;
			}
		}
	}
	if (v[1] == 1) {
		cout << 1 << endl; return 0;
	}
	if (v[12] == 1) {
		if (v[11] == 1) {
			cout << 1 << endl; return 0;
		}	
	}
	REP(i, 0, 13) {
		REP(j, 0, v[i]) {
			hoge.push_back(i);
		}
	}
	REP(i, 0, hoge.size()) {
		if (i % 2 == 1) {
			ans.push_back(24 - hoge[i]);
		}
		else {
			ans.push_back(hoge[i]);
		}
	}
	sort(all(ans));
	int diff = hoge[1];
	REP(i, 1, ans.size()) {
		diff = min(diff, ans[i] - ans[i - 1]);
	}
	cout << diff << endl;
	return 0;
}