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
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
#define rep(i,a,b) for (int i = (a); i < (b); i++)
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef complex<double> com;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const int inf = 2e9;



int main(){
	int n; cin >> n;
	vector<int> a(n);
	rep(i, 0, n) {
		cin >> a[i];
	}
	a.push_back(0);
	sort(a.begin(), a.end());
	rep(i, 0, n + 1) {
		if (i % 2) a[i] = 24 - a[i];
	}
	int ans = 24;
	rep(i, 0, n + 1)rep(j, i+1, n + 1) {
		int d = abs(a[i] - a[j]);
		ans = min({ ans,d,24 - d });
	}
	printf("%d", ans);
	return 0;
}
