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"
#include<unordered_map>
#include<unordered_set>
#pragma warning(disable:4996)
using namespace std;
bool same(vector<int>v) {
	return all_of(v.begin(), v.end(), [&](const int a) {
		return a == v[0]; });

}
int main() {
	int N; cin >> N;
	vector<long double>as;
	for (int i = 0; i < N; ++i) {
		int a; cin >> a;
		as.push_back(a);
	}
	sort(as.begin(), as.end());
	long double ans = 0;
	for (int d = 0; d < N; ++d) {
		for (int c = d+1; c < N; ++c) {
			long double bo = as[c] - as[d];
			long double si;
			swap(c, d);
			if (c == N - 2 && d == N - 1) {
				si = as[N - 3] + as[N - 4];
			}
			else if (c==N-3&&d == N - 1) {
				si = as[N - 2] + as[N - 4];
			}
			else if (c == N - 3 && d == N - 2) {
				si = as[N - 1] + as[N - 4];
			}
			else if (d == N - 1) {
				si = as[N - 2] + as[N - 3];
			}
			else if (d == N - 2) {
				si = as[N - 1] + as[N - 3];
			}
			else {
				si = as[N - 1] + as[N - 2];
			}
			ans = max(ans, si / bo);
			swap(c, d);
		}
	}
	cout <<fixed<<setprecision(10)<< ans << endl;
	return 0;
}