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 INCANT cin.tie(0), cout.tie(0), ios::sync_with_stdio(false), cout << fixed << setprecision(20);
#define int long long
#define gcd __gcd
#define all(x) (x).begin(), (x).end()
template<class T>
bool chmax(T& a, T b){return (a = max(a, b)) == b;}
template<class T>
bool chmin(T& a, T b){return (a = (a > b ? b : a)) == b;}
#define _overload(_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(...) _overload(__VA_ARGS__, repi, _rep)(__VA_ARGS__)
#define _rev(i, n) revi(i, n, 0)
#define revi(i, a, b) for(int i = (int)(a - 1); i >= (int)(b); i--)
#define rev(...) _overload(__VA_ARGS__, revi, _rev)(__VA_ARGS__)
#define each(i, n) for(auto&& i: n)
const int INF = 1e18, MOD = 1e9 + 7;
signed main() {
    INCANT;
    int n, tmp = 0;
    double a[1111], res = -INF;
    cin>>n;
    rep(i, n){
        cin>>a[i];
    }
    sort(a, a + n);
    rep(i, n){
        rep(j, i + 1, n){
            if(i == n - 2){
                chmax(res, (a[n - 3] + a[n - 4]) / (a[j] - a[i]));
            }else if(i == n - 3){
                if(j == n - 2){
                    chmax(res, (a[n - 1] + a[n - 4]) / (a[j] - a[i]));
                }else{
                    chmax(res, (a[n - 2] + a[n - 4]) / (a[j] - a[i]));
                }
            }else{
                chmax(res, (a[n - 1] + a[n - 2]) / (a[j] - a[i]));
            }
        }
    }
    cout<<res<<endl;
}

