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>
int n, a[200010], p, q;
double l, r, m, x;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  std::sort(a + 1, a + n + 1);
  l = 0;
  r = (a[n] - a[1] + 1) / 6.0;
  p = 1;
  q = n;
  while (r - l > 1e-7) {
    m = (l + r) / 2;
    if (a[p] - a[1] > 2 * m) {
      while (a[p] - a[1] > 2 * m) p--;
      p++;
    } else {
      while (p < n && a[p + 1] - a[1] - 2 * m < 1e-8) p++;
      p++;
    }
    if (a[n] - a[q] > 2 * m) {
      while (a[n] - a[q] > 2 * m) q++;
      q--;
    } else {
      while (q > 1 && a[n] - a[q - 1] - 2 * m < 1e-8) q--;
      q--;
    }
    if (p < q && a[q] - a[p] > 2 * m)
      l = m;
    else
      r = m;
  }
  m = (l + r) / 2;
  p = 1;
  x = a[1] + m * 2;
  while (p < n && a[p + 1] < x + 1e-4) p++;
  x = a[n] - m * 2;
  q = n;
  while (q > 1 && a[q - 1] > x - 1e-4) q--;
  p++;
  q--;
  if (p <= n && q >= p)
    x = (a[p] * 1.0 + a[q]) / 2;
  else
    x = 0;
  printf("%.6f\n%.6f %.6f %.6f\n", m, a[1] + m, x, a[n] - m);
}