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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>

#define int long long
#define MOD7 1000000007
#define MOD9 1000000009

#define rep(i, n) for (int i = 0; i < (n); i++)
#define itrep(i, a) for (auto i = (a).begin(); i != (a).end(); i++)
#define REP(i, a, n) for (int i = (a); i <= (n); i++)
#define all(a) (a).begin(), (a).end()
#define mp(a, b) make_pair((a), (b))

using namespace std;

int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, -1, 0, 1 };

template<class T> void inputVector(vector<T>& v, int n) {
    v.resize(n);
    for (int i = 0; i < v.size(); i++) cin >> v[i];
}

signed main() {
	int N;
	cin >> N;

	vector<int> D;
	inputVector(D, N);
	sort(all(D));

	multiset<int> st;
	st.insert(0);
 
	rep(i, N) {
		st.insert(D[i] * (i % 2 ? -1 : 1));
	}

	vector<int> v;
	for (int num : st) {
		v.push_back(num);
	}

	int ret = 100000;
	rep(i, N) {
		ret = min(ret, v[i + 1] - v[i]);
	}
	ret = min(ret, abs(v[0] + 24 - v[N]));

	cout << ret << endl;

    return 0;
}
