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 i, k, m, n;
  int a[300000];
  double b[10];
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  double h, l;
  h = 0;
  l = a[n];
  while ((l - h) > 1e-8) {
    double mid = (h + l) / 2.0;
    k = 1;
    double j = a[1] + 2 * mid;
    i = upper_bound(a, a + n, j) - a;
    while (i <= n) {
      if (i == n && a[i] + 1e-8 < j) break;
      k++;
      j = a[i] + 2 * mid;
      if (k > 3) break;
      i = upper_bound(a, a + n, j) - a;
    }
    if (k <= 3)
      l = mid;
    else
      h = mid;
  }
  double mid = l;
  k = 1;
  double j = a[1] + 2 * mid;
  b[k] = a[1] + mid;
  for (i = 1; i <= n; i++)
    if (a[i] > j + 1e-8) {
      k++;
      b[k] = a[i] + mid;
      j = a[i] + 2 * mid;
    }
  if (k == 1) b[2] = b[1], b[3] = b[2];
  if (k == 2) b[3] = b[2];
  printf("%.6lf\n", l);
  for (i = 1; i < 3; i++) printf("%.6lf ", b[i]);
  printf("%.6lf\n", b[i]);
}