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 <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>

using namespace std;

double calFormula(const int64_t &A, const int64_t &B, const int64_t &C, const int64_t &D)
{
    return (double)(A+B) / (C-D);
}

int main()
{

    int N;
    cin >> N;

    vector<int64_t> list(N);
    for(int i = 0; i < N; i++)
    {
        cin >> list[i];
    }

    sort(list.begin(),list.end());
    reverse(list.begin(),list.end());

    double max_val = 0;
    //loop

    for(int a = 0; a < N; a++)
    {
        if(a > 10){ break; }
        for(int b = a + 1; b < N; b++)
        {
            if(b > 10){ break; }
            for(int c = 0; c < N; c++)
            {
                if(a == c || b == c){ continue; }
                for(int d = c + 1; d < N; d++)
                {
                    if(a == d || b == d){ continue; }
                    max_val = max(max_val, calFormula( list[a], list[b], list[c], list[d] ));
                }
            }
        }
    }


    cout << fixed << setprecision(5);
    cout << max_val << endl;

    return 0;
}