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>
#include <stdio.h>
#include <math.h>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(), (v).end()
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define BIT(x,n) bitset<n>(x)

typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> PP;

struct edge {
	int to,cost;
	edge(int t,int c):to(t),cost(c) {}
};

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

//-----------------------------------------------------------------------------

int n;
int d[51],cnt[51][51];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin>>n;
	REP(i,n) cin>>d[i+1];

	sort(d,d+n+1);

	FOR(i,1,n+1) {
		if(!(i&1)) d[i]=24-d[i];
	}

	int mi=INF;
	FOR(i,1,n+1) {
		int x=abs(d[i]-d[i-1]);
		if(i==1) mi=d[i];
		else mi=min(min(mi,abs(d[i]-d[i-2])),min(x,24-x));
	}

	cout<<mi<<endl;

	return 0;
}
