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 rep(i,N) for(int (i)=0;(i) < (N); (i)++)
#define all(V) V.begin(), V.end()
using i64 = int_fast64_t;
using P = pair<int,int>;

int main(){
    int N;
    cin >> N;
    vector<int> n(N);
    rep(i,N) cin >> n[i];

    sort(all(n));
    vector<pair<int, P>> diff;
    for(int i=0;i<N;i++){
        for(int j=i+1;j<N;j++){
            diff.push_back({n[j]-n[i],{j,i}});
        }
    }

    double ans = 0;

    for(auto p : diff){ 
        int i = p.second.second, j = p.second.first, k = 0;
        int sum = 0, d = p.first;
        for(int m = N-1; m >= 0; m--){
            if(m != i && m != j) {
                sum += n[m];
                k = m;
                break;
            }
        }
        for(int m = N-1; m >= 0; m--){
            if(m != i && m != j && m != k) {
                sum += n[m];
                break;
            }
        }
        double a = (double)sum / (double)d ;
        ans = max(ans, a);
    } 

    cout << setprecision(15) << ans << endl;
}

