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 int long long
#define DBG 0
#define dump(o) if(DBG){cerr<<#o<<" "<<o<<endl;}
#define dumpc(o) if(DBG){cerr<<#o; for(auto &e:(o))cerr<<" "<<e;cerr<<endl;}
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define each(it,c) for(auto it=(c).begin();it!=(c).end();it++)
#define all(c) c.begin(),c.end()
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9 + 7);

signed main() {
	cout << fixed << setprecision(7);
	int N; cin >> N;
	vector<double> a(N); rep(i, 0, N) { cin >> a[i]; }
	sort(a.begin(), a.end());
	double ans = (a[0] + a[1]) / (a[2] - a[3]);
	rep(i, 1, a.size()) {
		double C = a[i];
		rrep(j, 0, i) {
			double D = a[j];
			double A, B;
			int cnt = 0;
			rrep(k, 0, a.size()) {
				if (cnt == 2)break;
				if (k == i || k == j)continue;
				if (cnt == 0)A = a[k];
				if (cnt == 1)B = a[k];
				cnt++;
			}
			dump(A);
			dump(B);
			dump(C);
			dump(D);
			ans = max(ans, (A + B) / (C - D));
		}
	}
	cout << ans << endl;
	return 0;
}