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 <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
    cout << fixed << setprecision(8);
    int n;
    cin >> n;
    int A[n + 1];
    for(int i = 0; i < n; i++) cin >> A[i];
    sort(A, A + n);
    double ans = -1;
    for(int i = 0; i < n; i++)
    {
        for(int j = i + 1; j < n; j++)
        {
            int last = n - 1;
            while(last == i || last == j) last--;
            int a = A[last];
            last--;
            while(last == i || last == j) last--;
            int b = A[last];
            double x = (a + b) / (double)(A[j] - A[i]);
            ans = max(x, ans);
        }
    }
    cout << ans << endl;
	return 0;
}


