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>
int read(){
	char c=getchar();while (c!='-'&&(c<'0'||c>'9'))c=getchar();
	int k=1,kk=0;if (c=='-')c=getchar(),k=-1;
	while (c>='0'&&c<='9')kk=kk*10+c-'0',c=getchar();return k*kk;
}using namespace std;
int n,mi,a[1000],ff,f[1000];
bool pd(int x){
	for (int i=1;i<=n;i++){
		ff=0;f[i]=a[i];
		for (int j=0;j<i;j++)
			if (min(max(f[i],f[j])-min(f[i],f[j]),min(f[i],f[j])-max(f[i],f[j])+24)<x)ff=1;
		if (ff)ff=0,f[i]=24-a[i];
		for (int j=0;j<i;j++)
			if (min(max(f[i],f[j])-min(f[i],f[j]),min(f[i],f[j])-max(f[i],f[j])+24)<x)ff=1;
		if (ff)return 0;
	}return 1;
}
int main(){
	n=read();mi=100;
	for (int i=1;i<=n;i++)a[i]=read();sort(a+1,a+1+n);
	for (int i=a[1];i>=0;i--)
		if (pd(i)){cout<<i<<endl;return 0;}
}
