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>

#define rep(i,n) for(int i=0; i<(n); i++)
#define reps(i,x,n) for(int i=x; i<(n); i++)
#define rrep(i,n) for(int i=(n)-1; i>=0; i--)
#define all(X) (X).begin(),(X).end()
#define X first
#define Y second
#define pb push_back
#define eb emplace_back

using namespace std;
typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

template<class A, size_t N, class T> void Fill(A (&a)[N], const T &v){ fill( (T*)a, (T*)(a+N), v ); }

const ll INF = 1e9+7;


int main(){
	//ios_base::sync_with_stdio(false);
	int N, ans=0;
	int D[200];

	cin >> N;
	rep(i,N) cin >> D[i];
	D[N] = 0;
	N++;
	sort(D, D+N);

	rep(i,N) if(i%2) D[i] *= -1;
	sort(D, D+N);
	rep(i,N) D[i+N] = D[i] + 24;

	ans = INF;
	rep(i,N*2-1) chmin(ans, D[i+1]-D[i]);

	cout << ans << endl;

	return 0;
}
