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;
const double eps = 0.000000001;
double x[300000];
double p[3];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%lf", &x[i]);
  sort(x, x + n);
  int j;
  bool q;
  double l = 0, r = 1000 * 1000 * 1000, m;
  while ((r - l - 0.000001) > eps) {
    m = (r + l) / 2;
    j = 0;
    q = false;
    p[0] = x[0] + m;
    for (int i = 1; i < n; i++)
      if (abs(p[j] - x[i]) - m > eps) {
        if (j == 2) {
          q = true;
          break;
        }
        j++;
        p[j] = x[i] + m;
      }
    if (q)
      l = m;
    else
      r = m;
  }
  m = ((double)trunc(r * 1000000)) / 1000000;
  j = 0;
  p[0] = x[0] + m;
  for (int i = 1; i < n; i++)
    if (abs(p[j] - x[i]) - m > eps) {
      j++;
      p[j] = x[i] + m;
    }
  while (j < 2) p[++j] = 0;
  printf("%.6lf\n", m);
  for (int i = 0; i < 3; i++) printf("%.6lf ", p[i]);
  printf("\n");
  return 0;
}