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;
int main() {
    int n;
    int A1 = 0;
    int B1 = 0;
    int C1 = 0;
    int D1 = 0;
    int A2 = 0;
    int B2 = 0;
    int C2 = 0;
    int D2 = 0;
    int enter;
    unsigned long long min = 1000000000000;
    int min_x;
    int min_y;
    int i, j;
    cin >> n;
    vector<int> ent;
    vector<int> a;
    vector<int> b;
    for(i = 0; i < n; i++) {
        cin >> enter;
        ent.push_back(enter);
    }
    sort(ent.begin(), ent.end(), greater<int>());
    for(i = 0; i < n; i++) {
        a.push_back(ent[i]);
        b.push_back(ent[i]);
    }
    for(i = 0; i < n; i++) {
        for(j = 0; j < n; j++) {
            if(a[i] - a[j] > 0) {
                if(a[i] - a[j] < min) {
                    min = a[i] - a[j];
                    min_x = i;
                    min_y = j;
                }
            }
        }
    }

    C1 = a[min_x];
    D1 = a[min_y];

    a.erase(a.begin() + min_x);

    a.erase(a.begin() + min_y - 1);

    A1 = a[0];
    B1 = a[1];

    A2 = b[0];
    B2 = b[1];

    b.erase(b.begin());
    b.erase(b.begin());

    min = 1000000000000;
    for(i = 0; i < n; i++) {
        for(j = 0; j < n; j++) {
            if(b[i] - b[j] > 0) {
                if(b[i] - b[j] < min) {
                    min = b[i] - b[j];
                    min_x = i;
                    min_y = j;
                }
            }
        }
    }
    C2 = b[min_x];
    D2 = b[min_y];

    if((A1 + B1) / (C1 - D1) > (A2 + B2) / (C2 - D2)) {
        printf("%16f\n", (double)(A1 + B1) / (C1 - D1));
    } else {
        printf("%16f\n", (double)(A2 + B2) / (C2 - D2));
    }

    return 0;
}
