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;
int main() {
  int N;
  scanf("%d", &N);
  int T[N];
  for (int i = 0; i < N; i++) scanf("%d", &T[i]);
  sort(T, T + N);
  N = unique(T, T + N) - T;
  int a = 0;
  int b = 1000000000;
  int R[3];
  while (a != b) {
    int k = 0;
    int s = (a + b) / 2;
    R[0] = R[1] = R[2] = -1;
    for (int i = 0; i < 3 && k < N; i++) {
      R[i] = k;
      int j = k + 1;
      while (j < N && T[j] - T[k] <= s) j++;
      k = j;
    }
    if (k < N)
      a = s + 1;
    else
      b = s;
  }
  int s = a;
  int k = 0;
  R[0] = R[1] = R[2] = -1;
  for (int i = 0; i < 3 && k < N; i++) {
    R[i] = k;
    int j = k + 1;
    while (j < N && T[j] - T[k] <= s) j++;
    k = j;
  }
  double d = (double)a / 2;
  printf("%0.6lf\n", d);
  for (int i = 0; i < 3; i++) {
    printf("%0.6lf ", T[R[i] == -1 ? N - 1 : R[i]] + d);
  }
  printf("\n");
}