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"
#define int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define inv(i, a, b) for (int i = (int)(a); i >= (int)(b); i--)
#define ALL(obj) begin(obj), end(obj)
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))
#define LOWER_BOUND(A, key) distance(A.begin(), lower_bound(ALL(A), key))
#define UPPER_BOUND(A, key) distance(A.begin(), upper_bound(ALL(A), key))
#define endl "\n"
#define Endl "\n"

using namespace std;
const int MOD = (int)(1e9 + 7);
const int INF = (int)(1e13 + 7);

int CEIL(int a, int b) { return (a >= 0 ? (a + (b - 1)) / b : (a - (b - 1)) / b); } //ceil() for int
int mod(int a, int b) { return a >= 0 ? a % b : a - (b * CEIL(a, b)); }             //always return positive num
int bpm(int x, unsigned int y) {                                                    //return x^y in order(log(y))
  if (x == 0) return 0;
  if (y == 0) return 1;
  int ans   = 1;
  int digit = (int)((log((double)y) / log((double)2) / 1 + 1));
  x %= MOD;
  for (int i = 0; i < digit; i++) {
    if (((y >> i) & 1u) == 1) ans = ans * x % MOD;
    x = x * x % MOD;
  }
  return ans;
}
template<class T>
void cumulative_sum(T &container) {
  for (int i = 0; i < container.size() - 1; i++) container[i + 1] += container[i];
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<double> a(N);
  rep(i, N) cin >> a[i];
  double ans = -INF;
  sort(ALL(a));
  vector<vector<double>> check(N * (N - 1) / 2, vector<double>(3));
  int now = 0;
  rep(i, N - 1) {
    FOR(j, i + 1, N) {
      check[now++] = {a[j] - a[i], a[i], a[j]};
    }
  }
  sort(ALL(check));
  rep(i, N - 1) {
    FOR(j, i + 1, N) {
      rep(k, check.size()) {
        if (a[i] != check[k][1] and a[i] != check[k][2] and a[j] != check[k][1] and a[j] != check[k][2]) {
          ans = max(ans, (a[i] + a[j]) / check[k][0]);
          break;
        }
      }
    }
  }
  cout << fixed << setprecision(10) << ans << endl;
}
