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>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define fi first
#define se second
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, pii> pip;
const ll INF = 1e9+100;
const ll MOD = 1e9+7;
const double EPS  = 1e-10;
const bool debug = 0;
//---------------------------//

int N,A,B,C,D;
vector<int> a;
double ans = -INF;

int main(){
	cin >> N;
	a.resize(N);
  REP(i,N)cin >> a[i];
  sort(a.begin(),a.end());
  
  REP(i, N) {
  	FOR(j, i + 1, N) {
  		C = a[j]; D= a[i];
  		
  		int p = N - 1;
  		while (i == p || j == p) p--;
  		A = a[p]; p--;
  		
  		while (i == p || j == p) p--;
  		B = a[p];
  		
  		chmax(ans, (double)(A+B)/(C-D));
		}
  }
  
  printf("%.10f\n",ans);
	return 0;
}