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<algorithm>
#include<vector>
#include<string>
#include<set>
#include<queue>
#include<stack>
#include<bitset>
#include<functional>
#include<map>
#include<unordered_set>
using namespace std;
//int p = 998244353;
int p = 1000000007;
#define int long long
#define vel vector<long long>
#define vvel vector<vel>
#define rep(i,n) for(int i=0;i<n;i++)
#define sor(v) sort(v.begin(),v.end())
#define mmax(a,b) a=max(a,b)
#define mmin(a,b) a=min(a,b)
#define mkp make_pair
#define pin pair<int,int>
#define qin pair<int,pin>
#define V vector
#define Endl endl
#define veb vector<bool>
#define rev(s) reverse(s.begin(),s.end())
int max_kai = 150000;
vel kai(max_kai, 1);
vel inv_kai;
int rui(int a, int n) {
	if (n == 0) { return 1; }
	int x = rui(a, n / 2);
	x *= x; x %= p;
	if (n % 2 == 1) { x *= a; x %= p; }
	return x;
}
int inv(int a) { return rui(a, p - 2); }
void make_kai() {
	for (int i = 1; i < max_kai; i++) { kai[i] = kai[i - 1] * i; kai[i] %= p; }
	inv_kai = kai;
	rep(i, max_kai) { inv_kai[i] = inv(kai[i]); }
}
int nCr(int n, int r) {
	int a = kai[n] * inv_kai[r]; a %= p;
	a *= inv_kai[n - r]; a %= p;
	return a;
}
int inf = 6000000000000;
vel dijk(vvel &way, vel &cost, int s) {
	int n = way.size(); int m = 7 * n;
	vel ans(m, inf);
	priority_queue<pin, vector<pin>, greater<pin>> que;//cost,pla
	que.push(mkp(0,s));
	while (!que.empty()) {
		pin x = que.top(); que.pop();
		int pla = x.second;
		if (ans[pla] == inf) {
			int sum_cost = x.first;
			ans[pla] = sum_cost;
			int nex_cost = sum_cost + cost[pla];
			int mod = nex_cost % 7;
			for (auto x : way[pla/7]) {
				que.push(mkp(nex_cost, x * 7 + mod));
			}
		}
	}
	return ans;
}
vel pa;
int root(int x) {
	if (pa[x] == -1) { return x; }
	int ans = root(pa[x]); pa[x] = ans;
	return ans;
}
void marge(int x, int y) {
	x = root(x);
	y = root(y);
	if (x != y) {pa[x] = y;}
}
#define lower(h,val) lower_bound(h.begin(),h.end(),val)-h.begin()
#define upper(h,val) upper_bound(h.begin(),h.end(),val)-h.begin()
int sol(int a, int b, int c, int d) {
	return abs(a*c - b * d);
}
signed main(){
	int n; cin >> n;
	V<double> a(n);
	rep(i, n) { cin >> a[i]; }
	sor(a);
	double ans=0;
	rep(i, n) {
		rep(j, i) {
			vel ex;
			for (int k = n-1; k >= n-4; k--) {
				if (k != i && k != j) {
					ex.push_back(k);
				}
			}
			mmax(ans, (a[ex[0]] + a[ex[1]]) / (a[i] - a[j]) );
		}
	}
	printf("%.6lf\n", ans);
	return 0;
}
