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

#define INF (1 << 29)

int main(){
  int n;
  vector<int> a;
  int temp;
  scanf("%d",&n);
  for(int i = 0;i < n;++i){
    scanf("%d",&temp);
    a.push_back(temp);
  }
  sort(a.begin(), a.end());
  double ans = 0;
  for(int i = 0;i <= n - 4;++i){
    ans = max(ans, (double)((double)(a[n - 1] + a[n - 2]) / (double)(a[i + 1] - a[i])));
  }
  ans = max(ans, (double)((double)(a[n - 3] + a[n - 4]) / (double)(a[n - 1] - a[n - 2])));
  ans = max(ans, (double)((double)(a[n - 1] + a[n - 4]) / (double)(a[n - 2] - a[n - 3])));
  printf("%.6lf\n",ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;

#define INF (1 << 29)

int main(){
  int n;
  vector<int> a;
  int temp;
  scanf("%d",&n);
  for(int i = 0;i < n;++i){
    scanf("%d",&temp);
    a.push_back(temp);
  }
  sort(a.begin(), a.end());
  double ans = 0;
  for(int i = 0;i <= n - 4;++i){
    ans = max(ans, (double)((double)(a[n - 1] + a[n - 2]) / (double)(a[i + 1] - a[i])));
  }
  ans = max(ans, (double)((double)(a[n - 3] + a[n - 4]) / (double)(a[n - 1] - a[n - 2])));
  ans = max(ans, (double)((double)(a[n - 1] + a[n - 4]) / (double)(a[n - 2] - a[n - 3])));
  printf("%.6lf\n",ans);
  return 0;
}

```