#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    multiset<int, greater<int>> s;
    for(int i = 0; i < n; ++i) {
        cin >> a[i];
        s.insert(a[i]);
    }

    double ans = 0;
    for(int i = 0; i < n; ++i) {
        for(int j = i + 1; j < n; ++j) {
            s.erase(a[i]);
            s.erase(a[j]);
            double x = abs(a[i] - a[j]);
            double y = *s.begin() + *(next(s.begin()));
            s.insert(a[i]);
            s.insert(a[j]);
            ans = max(ans, y / x);
        }
    }

    cout << fixed << setprecision(10) << ans << endl;
}


### Prompt

Your challenge is to write a cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    multiset<int, greater<int>> s;
    for(int i = 0; i < n; ++i) {
        cin >> a[i];
        s.insert(a[i]);
    }

    double ans = 0;
    for(int i = 0; i < n; ++i) {
        for(int j = i + 1; j < n; ++j) {
            s.erase(a[i]);
            s.erase(a[j]);
            double x = abs(a[i] - a[j]);
            double y = *s.begin() + *(next(s.begin()));
            s.insert(a[i]);
            s.insert(a[j]);
            ans = max(ans, y / x);
        }
    }

    cout << fixed << setprecision(10) << ans << endl;
}


```