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.util.*;
import java.io.*;

// Solution

public class Main 
{    
  public static void main (String[] argv)
  {
	  new Main();
  } 
     
  
  boolean test = false;
  
  int n;
  Integer[] a;
  final int nS = 3;
  
  public Main() {
	  FastReader in = new FastReader(new BufferedReader(new InputStreamReader(System.in)));
	  //FastReader in = new FastReader(new BufferedReader(new FileReader("Main.in")));
      
      n = in.nextInt();
      a = new Integer[n];
      for(int i = 0; i < n; ++i)a[i] = 2 * in.nextInt();
      Arrays.sort(a);
      
      int lo = 0, hi = a[n-1] - a[0];
      while (lo <= hi) {
          int mid = lo + (hi - lo) / 2;
          if (canCover(mid)) hi = mid - 1;
          else               lo = mid + 1;
      }
      
      
      System.out.println(format(lo));
      int[] centers = findCenters(lo);
      System.out.println(format(centers[0]) + " " + format(centers[1]) + " " + format(centers[2]));
  }
  
  private boolean canCover(int d) {
      for (int i = 0, s = 0; i < nS; ++i) {
          long ex = a[s] + 2L * d;
          if (ex >= a[n-1]) return true;
          
          s = bstSearch((int)ex);
          if (s == n) return true;
      }
      return false;
  }
  
  private int[] findCenters(int d) {
      int[] centers = new int[nS];
      for (int i = 0, s = 0; i < nS; ++i) {
          centers[i] = a[s] + d;
          
          long ex = a[s] + 2L * d;
          if (ex >= a[n-1]) break;
          
          s = bstSearch((int)ex);
          if (s == n) break;
      }
      return centers;
  }
  
  private int bstSearch(int val) {
      int lo = 0, hi = n - 1;
      if (val >= a[hi]) return n;
      
      while (lo <= hi) {
          int mid = lo + (hi - lo)  /2;
          if (a[mid] <= val) lo = mid + 1;
          else hi = mid - 1;
      }
      return lo;
  }
  
  private String format(int v) {
      if (v % 2 == 0) return "" + (v/2) + ".000000";
      return "" + (v/2) + ".500000";
  }
  
  private int max(int a, int b) {
      return a >  b ? a : b;
  }
  
  private int min(int a, int b) {
      return a >  b ? b : a;
  }
  
  
  static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader(BufferedReader in)
        {            
            br = in;
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    String line = br.readLine();
                    if (line == null || line.length() == 0) return "";
                    st = new StringTokenizer(line);
                }
                catch (IOException  e)
                {
                    return "";
                    //e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt()
        {
            return Integer.parseInt(next());
        }
 
        long nextLong()
        {
            return Long.parseLong(next());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                return "";
                //e.printStackTrace();
            }
            return str;
        }
    }
}