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 <sstream>
#include <algorithm>
#include <vector>
#include <utility>
#include <functional>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <deque>
#include <ctime>
  
using namespace std;
  
#define rep(i,n) REP(i,0,n)
#define REP(i,s,e) for(int i=(s); i<(int)(e); i++)
#define pb push_back
#define mp make_pair
#define all(r) (r).begin(),(r).end()
#define rall(r) (r).rbegin(),(r).rend()
#define fi first
#define se second
#define println(X) cout<<X<<endl;
#define DBG(X) cout<<#X<<" : "<<X<<endl;
  
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
 
const int INF = 1e9;
const ll LINF = 1e18;
const ll MOD = 1e9 + 7;
double EPS = 1e-8;
const double PI = acos(-1);

int main(){
	int n;
	cin>>n;
	vi v(n);
	rep(i, n) cin>>v[i];
	sort(all(v));
	double ans = 0.0;
	for(int i = 0; i < n; i++) {
		for(int j = i + 1; j < n; j++) {
			int c = v[j], d = v[i];
			int a, b;
			for(int k = n - 1; k >= 0; k--) if(v[k] != c && v[k] != d){
				a = v[k];
				break;
			}
			for(int k = n - 1; k >= 0; k--) if(v[k] != c && v[k] != d && v[k] != a){
				b = v[k];
				break;
			}
			ans = max(ans, (double)(a+b)/(c-d));
		}
	}
	printf("%.15lf\n", ans);
}