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>
int x[2 * 100001];
using namespace std;
double counter, p, d;
int n;
bool isValid(double d) {
  double p = d + x[0];
  counter = 1;
  for (int i = 0; i < n; i++) {
    if (x[i] > (p + d)) {
      p = (double)x[i] + d;
      counter++;
    }
  }
  if (counter > 3) {
    return false;
  }
  return true;
}
void getSolution() {
  double p = d + x[0];
  counter = 1;
  printf("%0.6f\n", d);
  printf("%0.6f ", p);
  for (int i = 0; i < n; i++) {
    if (x[i] > (p + d) + 0.000001) {
      p = (double)x[i] + d;
      printf("%0.6f ", p);
      counter++;
    }
    if (counter == 3) break;
  }
  while (counter < 3) {
    counter++;
    printf("%0.6f ", p);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x[i]);
  }
  sort(x, x + n);
  double e = 10000000000;
  double s = 0;
  while (e - s > 0.0000001) {
    d = e + (s - e) / 2;
    if (isValid(d)) {
      e = d;
    } else {
      s = d;
    }
  }
  getSolution();
  return 0;
}