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

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i=0;i<n;i++){
        cin >> a[i];
    }
    sort(a.begin(),a.end());
    double ans = 0.0;
    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            int x = n-1, y = n-2;
            while(x==i || x==j) x--;
            while(y==i || y==j || y==x) y--;
            ans = max(ans, double(a[x]+a[y])/(a[j]-a[i]));
        }
    }
    printf("%.8lf\n",ans);
    return 0;
}


### Prompt

Please formulate 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);
    for(int i=0;i<n;i++){
        cin >> a[i];
    }
    sort(a.begin(),a.end());
    double ans = 0.0;
    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            int x = n-1, y = n-2;
            while(x==i || x==j) x--;
            while(y==i || y==j || y==x) y--;
            ans = max(ans, double(a[x]+a[y])/(a[j]-a[i]));
        }
    }
    printf("%.8lf\n",ans);
    return 0;
}


```