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>
#define rep(i,n) for(int i=0;i<n;++i)
#define rrep(i,n) for(int i=n-1;i>=0;--i)
#define yesno(flg) if(flg){cout<<"yes"<<endl;}else{cout<<"no"<<endl;}
#define vi vector<int>
#define MAX_N 200005
#define i197 1000000007

using namespace std;


typedef long long ll;
typedef pair<ll,ll> P1;
typedef pair<int,P1> P2;
const int inf=1000000000;
P1 xy[MAX_N]={};

int main() {
	int n;
	cin>>n;
	int a[1001];
	rep(i,n)cin>>a[i];
	sort(a,a+n);
	int k=1000000000;
	rep(i,n-3){
		if(a[i+1]-a[i]<k){
			k=a[i+1]-a[i];
		}
	}
	double r=((double)a[n-1]+(double)a[n-2])/(double)k;
	r=max(r,((double)a[n-1]+(double)a[n-4])/((double)a[n-2]-(double)a[n-3]));
	r=max(r,((double)a[n-3]+(double)a[n-4])/((double)a[n-1]-(double)a[n-2]));
	printf("%.5f",r);
	return 0;
}

