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 <algorithm>
#include <bitset>
#include <cfloat>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits.h>
#include <map>
#include <math.h>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <string.h>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
#define int long long
#define ll long long
#define eps LDBL_EPSILON
#define mod 1000000007
#define int long long
#define double long double
#define INF LLONG_MAX/10
#define P pair<int,int>
#define prique priority_queue
using namespace std;
int n, d[55];
signed main() {
	cin >> n;
	rep(i, n)cin >> d[i];
	d[n] = 0;
	sort(d, d + n + 1);
	int a = 0, b = 0, ans = 24;
	REP(i, n) {
		if (i % 2) {
			ans = min(ans, d[i] - b);
			ans = min(ans, 24 - a - d[i]);
			b = d[i];
		}
		else {
			ans = min(ans, d[i] - a);
			ans = min(ans, 24 - b - d[i]);
			a = d[i];
		}
	}
	cout << ans << endl;
	return 0;
}