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;

using i64 = long long int;

#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(a) a.begin(), a.end()

struct IoSetup {
    IoSetup() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(10);
        cerr << fixed << setprecision(10);
    }
} iosetup;

int main(){
    i64 n;
    cin >> n;
    vector<double> a(n);
    rep(i, n) cin >> a[i];
    sort(all(a));
    double ans = 0;
    rep(i, n - 1){
        rep(j, i + 1, n){
            double cd = a[j] - a[i];
            double ab = 0;
            i64 cnt = 0;
            for(i64 k = n - 1; k >= 0; --k){
                if(cnt == 2) break;
                if(k != i && k != j){
                    ab += a[k];
                    ++cnt;
                }
            }
            ans = max(ans, ab / cd);
        }
    }
    cout << ans << endl;
}
