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 JAVA):

Solution:
import java.io.File;
import java.io.PrintStream;
import java.util.Collections;
import java.util.Scanner;
import java.util.Vector;


public class ProblemC {
  public static void main(String[] args) {
    Scanner s = null;
    s = new Scanner(System.in);
//    try {s = new Scanner(new File("c.in"));}catch(Exception e) {e.printStackTrace();}
    
    double d=0;
    double x1=0, x2=0, x3=0;
    int n = s.nextInt();
    Vector<Double> v = new Vector<Double>();
    
    for (int i=0; i<n; i++) {
      v.add(s.nextDouble());
    }
    
    if (n == 0) {
      d=0; x1=0; x2=0; x3=0;
    }
    else if (n==1) {
      d=0; x1=v.get(0); x2=v.get(0); x3=v.get(0);
    }
    else if (n==2) {
      d=0; x1=v.get(0); x2=v.get(0); x3=v.get(1);
    }
    else if (n==3) {
      d=0; x1=v.get(0); x2=v.get(1); x3=v.get(2);
    }
    else {
      Collections.sort(v);
      
      int p1 = 0;
      int p2 = n-1;
      double min = 2000000001;
      while (true) {
        double l1 = v.get(p1) - v.get(0);
        double l2 = v.get(n-1) - v.get(p2);
        double l3;
        if (p1+1 > p2-1) 
          l3 = 0;
        else 
          l3 = v.get(p2-1) - v.get(p1+1);
        
        double nec = Math.max(Math.max(l1, l2), l3);
//        System.out.println(p1 + " " + p2 + "|  " + l1 + " " + l3 + " " + l2 + ": " + nec);

        if (nec < min) {
          min = nec;
          d = nec / 2;
          x1 = (v.get(p1) + v.get(0)) / 2;
          if (p1+1 > p2-1) 
            x2 = 0;
          else
            x2 = (v.get(p1+1) + v.get(p2-1)) / 2;
          x3 = (v.get(p2) + v.get(n-1)) / 2;
        }        
        if (p1 == n-1 && p2 == 0) 
          break;
        else
        if (p1 == n-1) 
          p2--;
        else if (p2 == 0) 
          p1++;
        else if (v.get(p1+1) - v.get(0) < v.get(n-1) - v.get(p2-1))
          p1++;
        else 
          p2--;
        if (p1 > p2) break;
      }
    }
    
    System.out.println(d);
    System.out.println(String.format("%.6f", x1)+ " " +  String.format("%.6f", x2)+ " " +  String.format("%.6f", x3));
  }
}