Problem: The New Vasjuki village is stretched along the motorway and that's why every house on it is characterized by its shift relative to some fixed point — the xi coordinate. The village consists of n houses, the i-th house is located in the point with coordinates of xi.

TELE3, a cellular communication provider planned to locate three base stations so as to provide every house in the village with cellular communication. The base station having power d located in the point t provides with communication all the houses on the segment [t - d, t + d] (including boundaries).

To simplify the integration (and simply not to mix anything up) all the three stations are planned to possess the equal power of d. Which minimal value of d is enough to provide all the houses in the village with cellular communication.

Input

The first line contains an integer n (1 ≤ n ≤ 2·105) which represents the number of houses in the village. The second line contains the coordinates of houses — the sequence x1, x2, ..., xn of integer numbers (1 ≤ xi ≤ 109). It is possible that two or more houses are located on one point. The coordinates are given in a arbitrary order.

Output

Print the required minimal power d. In the second line print three numbers — the possible coordinates of the base stations' location. Print the coordinates with 6 digits after the decimal point. The positions of the stations can be any from 0 to 2·109 inclusively. It is accepted for the base stations to have matching coordinates. If there are many solutions, print any of them.

Examples

Input

4
1 2 3 4


Output

0.500000
1.500000 2.500000 3.500000


Input

3
10 20 30


Output

0
10.000000 20.000000 30.000000


Input

5
10003 10004 10001 10002 1


Output

0.500000
1.000000 10001.500000 10003.500000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int x[200005];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> x[i];
  sort(x + 1, x + 1 + n);
  double l = 0, r = 1e9;
  double E = 1e-9;
  double st[5];
  double ans[5];
  x[n + 1] = 2e9 + 3;
  while (l + E <= r) {
    double mid = (l + r) / 2;
    int pos = 1;
    int cnt = 0;
    while (pos <= n && cnt <= 2) {
      cnt++;
      st[cnt] = x[pos] + mid;
      while (abs(st[cnt] - x[pos]) <= mid + E) {
        pos++;
      }
    }
    if (pos <= n)
      l = mid + E;
    else {
      r = mid;
      for (int i = 1; i <= 3; i++) ans[i] = st[i];
    }
  }
  cout << fixed << setprecision(6) << l << "\n";
  for (int i = 1; i <= 3; i++) {
    cout << fixed << setprecision(6) << ans[i] << " ";
  }
}