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
#define _CRT_SECURE_NO_WARNINGS
#include "bits/stdc++.h"
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define FOR(i, m, n) for(int i=(m); i<(n); ++i)
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mp make_pair
#define pb push_back
#define Cout(x) cout << (x) << endl
#define Cout2(x, y) cout << (x) << " " << (y) << endl
#define dump(x) cerr << #x << " = " << (x) << endl;
typedef long long LL;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<string> VS;
typedef vector<bool> VB;
typedef vector<vector<int>> VVI;
typedef pair<int, int> PII;
const int inf = 1e9;
const double pi = acos(-1.0);

int main() {
	int n; cin >> n;
	VI a(n); rep(i, n)cin >> a[i];
	sort(all(a));
	
	double ans = -(double)inf;
	rep(i, n) {
		FOR(j, i + 1, n) {
			int cd = a[j] - a[i];
			VI v_ab;
			for (int k = n - 1; k >= 0; --k) {
				if (k != i && k != j)v_ab.pb(a[k]);
				if (sz(v_ab) == 2)break;
			}
			int ab = v_ab[0] + v_ab[1];

			ans = max(ans, (double)ab / cd);
		}
	}
	printf("%.20f\n", ans);
}
