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.*;


public class C
{
   public static void main(String[] args)
   {
      new C(new Scanner(System.in));
   }

   public C(Scanner in)
   {
      int N = in.nextInt();
      int[] vs = new int[N];
      for (int i=0; i<N; i++)
         vs[i] = in.nextInt();
      Arrays.sort(vs);

      double lo = 0;
      double hi = vs[vs.length-1]-vs[0];
      for (int u=0; u<200; u++)
      {
         if ((hi-lo) < 1e-8) break;
         
         double m = (hi+lo)*0.5;
         
         int cnt = 1;
         int last = 0;
         for (int i=1; i<vs.length; i++)
         {
            int d = vs[i]-vs[last];
            if (d > m)
            {
               last = i;
               cnt++;
            }
         }

         if (cnt > 3)
            lo = m;
         else
            hi = m;
      }

      double m = hi;
      double[] res = new double[3];

      int cnt = 0;
      int last = 0;
      for (int i=1; i<vs.length; i++)
      {
         int d = vs[i]-vs[last];
         if (d > m)
         {
            res[cnt++] = (vs[i-1]+vs[last])*0.5;
            last = i;
         }
      }

      res[cnt++] = (vs[vs.length-1]+vs[last])*0.5;

      m*=0.5;
      m+=1e-9;
      Arrays.sort(res);
      for (int i=0; i<3; i++) res[i]+=1e-9;
      System.out.printf("%.6f%n", m);
      System.out.printf("%.6f %.6f %.6f%n", res[0], res[1], res[2]);
   }
}