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 <array>
# include <cassert>
#include <cctype>
#include <climits>
#include <numeric>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <tuple>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <bitset>
# include <complex>
# include <chrono>
# include <random>
# include <limits.h>
# include <unordered_map>
# include <unordered_set>
# include <deque>
# include <cstdio>
# include <cstring>
#include <stdio.h>
#include<time.h>
#include <stdlib.h>
#include <cstdint>
#include <cfenv>

//#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
long long MOD = 1000000000 + 7;
constexpr long long INF = numeric_limits<LL>::max();
const double PI = acos(-1);
#define fir first
#define sec second
#define thi third
#define debug(x) cerr<<#x<<": "<<x<<'\n'
typedef pair<LL, LL> Pll;
typedef pair<LL, pair<LL, LL>> Ppll;
typedef pair<LL, pair<LL, bitset<100001>>> Pbll;
typedef pair<LL, pair<LL, vector<LL>>> Pvll;
typedef pair<LL, LL> Vec2;
struct Tll { LL first, second, third; };
struct Fll { LL first, second, third, fourd; };
typedef pair<LL, Tll> Ptll;
#define rep(i,rept) for(LL i=0;i<rept;i++)
#define Mfor(i,mf) for(LL i=mf-1;i>=0;i--)
int h, w, n, m, k, t, s, q, last;
double cnt, sum[10000], ans, dp[10000][10000], a[10000], b[10000];
string str, ss;
bool f;
char c;
int di[4][2] = { { 0,1 },{ 1,0 },{ -1,0 } ,{ 0,-1 } };
struct Edge { LL to, cost; };
vector<LL>vec[100100];
vector<Pll>v[200000];
map<Pll, LL>ma;
set<LL>st;
void YN(bool f) {
	if (f)
		cout << "YES" << endl;
	else
		cout << "NO" << endl;
}
void yn(bool f) {
	if (f)
		cout << "Yes" << endl;
	else
		cout << "No" << endl;
}

int main() {
	cin >> n;
	rep(i, n) {
		cin >> a[i];
	}
	sort(a, a + n);
	rep(i, n - 3) {
		if (i + 1 < n - 2)
			ans = max(ans, (a[n - 1] + a[n - 2]) / abs(a[i + 1] - a[i]));
	}
	ans = max(ans, (a[n - 1] + a[n - 4]) / abs(a[n - 2] - a[n - 3]));
	ans = max(ans, (a[n - 3] + a[n - 4]) / abs(a[n - 1] - a[n - 2]));
	printf("%.15f\n",ans);
	return 0;
}

