N different natural numbers are given. If you select four different ones and set them as $ A $, $ B $, $ C $, $ D $, the following formula

$ \ Frac {A + B} {C --D} $



I want to find the maximum value of.




Given N different natural numbers, choose 4 different from them and create a program to find the maximum value of the above formula.



Input

The input is given in the following format.


N
a1 a2 ... aN


The number N (4 ≤ N ≤ 1000) of natural numbers is given in the first line. The value ai (1 ≤ ai ≤ 108) of each natural number is given in the second line. However, the same natural number does not appear more than once (ai ≠ aj for i ≠ j).

Output

Outputs the maximum value of the above formula as a real number for a given N natural numbers. However, the error must not exceed plus or minus 10-5.

Examples

Input

10
1 2 3 4 5 6 7 8 9 10


Output

19.00000


Input

5
22 100 42 3 86


Output

9.78947


Input

6
15 21 36 10 34 5


Output

18.00000


Input

4
100000 99999 8 1


Output

28571.285714
#include<bits/stdc++.h>
using namespace std;

#define INF (1<<30)

int main(){
	int n;	cin>>n;
	vector<double> a(n);	for(int i=0;i<n;i++)cin>>a[i];
	sort(a.begin(), a.end());

	//{{{
	double used[2];
	double m=INF;
	for(int i=0;i<n;i++){
		for(int j=i+1;j<n;j++){
			if(a[j]-a[i]<m){
				m=a[j]-a[i];
				used[0]=a[i];	used[1]=a[j];
			}
		}
	}

	double ans1=0;
	for(int i=0;i<n;i++){
		for(int j=i+1;j<n;j++){
			bool flag=true;
			for(int k=0;k<2;k++){
				if(a[i]==used[k] || a[j]==used[k]){
					flag=false;
					break;
				}
			}
			if(flag==false)continue;

			double tmp=(a[i]+a[j])/m;
			ans1=max(ans1,tmp);
		}
	}
	//}}}
	double ans2=0;
	for(int i=0;i<n-2;i++){
		for(int j=i+1;j<n-2;j++){
			double tmp=(a[n-2]+a[n-1])/(a[j]-a[i]);
			ans2=max(ans2,tmp);
		}
	}

	double ans=max(ans1,ans2);

	printf("%.6lf\n",ans);

	return 0;

}

