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 main() {
  int n;
  scanf("%d", &n);
  int x[n];
  for (int i = 0; i < n; i++) scanf("%d", &x[i]);
  sort(x, x + n);
  cout << fixed;
  cout << setprecision(6);
  if (n == 1) {
    cout << "0\n";
    cout << (double)x[0] << " " << (double)x[0] << " " << (double)x[0] << "\n";
    return 0;
  }
  if (n == 2) {
    cout << "0\n";
    cout << (double)x[0] << " " << (double)x[1] << " " << (double)x[1] << "\n";
    return 0;
  }
  if (n == 3) {
    cout << "0\n";
    double x1 = x[0];
    double x2 = x[1];
    double x3 = x[2];
    cout << x1 << " " << x2 << " " << x3 << "\n";
    return 0;
  }
  double lo = 0;
  double hi = (x[n - 1] - x[0]) / 2;
  for (int i = 0; i < 100; i++) {
    double mid = (lo + hi) / 2.0;
    int j, k;
    for (j = 0, k = 0; j < n;) {
      if (k == 3) break;
      int start = x[j];
      int end = (int)((double)start + 2.0 * mid);
      while (x[j] <= end) {
        j++;
        if (j == n) break;
      }
      k++;
    }
    if (j == n) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  double mid = (lo + hi) / 2;
  cout << mid << "\n";
  int j;
  vector<double> c;
  for (j = 0; j < n;) {
    int start = x[j];
    int end = (int)((double)start + 2.0 * mid);
    c.push_back(start + mid);
    while (x[j] <= end) {
      j++;
      if (j == n) break;
    }
  }
  if (c.size() == 1)
    cout << c[0] << " " << c[0] << " " << c[0] << "\n";
  else if (c.size() == 2)
    cout << c[0] << " " << c[1] << " " << c[1] << "\n";
  else
    cout << c[0] << " " << c[1] << " " << c[2] << "\n";
  return 0;
}