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 V vector
#define VI vector<int>
#define VVI vector<vector<int>>
#define REP(i,n) for(int i=0;i<(n);i++)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define FOR(i,m,n) for(int i=m;i<n;i++)
#define MOD 1000000007
#define INF 1e9
#define F first
#define S second
#define All(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;

int main(void){
    int N;
    cin>>N;
    double ans=0.0;
    VI a(N);
    REP(i,N)cin>>a[i];
    sort(All(a));
    int m1=a[N-1],m2=a[N-2],m3=a[N-3],m4=a[N-4];

    FOR(i,1,N-2){
        int C=a[i],D=a[i-1];
        ans=max(ans,1.0*(m1+m2)/(C-D));
    }
    ans=max(ans,1.0*(m1+m4)/(a[N-2]-a[N-3]));
    ans=max(ans,1.0*(m3+m4)/(a[N-1]-a[N-2]));
    printf("%lf\n",ans);
}
