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;
using namespace std;
int h[200005];
int n;
bool ok(double power) {
  int x1 = (int)(upper_bound(h, h + n, h[0] + 2.0 * power + 1e-8) - h);
  if (x1 == n) {
    return 1;
  } else {
    int x2 = (int)(upper_bound(h + x1, h + n, h[x1] + 2.0 * power + 1e-8) - h);
    if (x2 == n) {
      return 1;
    } else {
      int x3 =
          (int)(upper_bound(h + x2, h + n, h[x2] + 2.0 * power + 1e-8) - h);
      if (x3 == n) return 1;
    }
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  int minval = (1 << 28);
  int maxval = -1;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &h[i]);
    minval = min(minval, h[i]);
    maxval = max(maxval, h[i]);
  }
  sort(h, h + n);
  double lo = 0.0, hi, mid;
  hi = 1e10;
  while (lo + 1e-7 < hi) {
    mid = (lo + hi) / 2.0;
    if (ok(mid)) {
      hi = mid;
    } else
      lo = mid;
  }
  double pos1, pos2, pos3;
  char g[100];
  sprintf(g, "%.6f", mid);
  string temp = g;
  stringstream q;
  q << temp;
  q >> mid;
  int x1 = (int)(upper_bound(h, h + n, h[0] + 2.0 * mid + 1e-7) - h);
  pos1 = (h[x1 - 1] + h[0]) / 2.0;
  if (x1 == n) {
    pos2 = pos3 = pos1;
  } else {
    int x2 = (int)(upper_bound(h + x1, h + n, h[x1] + 2.0 * mid + 1e-7) - h);
    pos2 = (h[x2 - 1] + h[x1]) / 2.0;
    if (x2 == n) {
      pos3 = pos2;
    } else {
      int x3 = (int)(upper_bound(h + x2, h + n, h[x2] + 2.0 * mid + 1e-7) - h);
      pos3 = ((long long)h[x3 - 1] + h[x2]) / 2.0;
    }
  }
  printf("%.6f\n", mid);
  printf("%.6f %.6f %.6f", pos1, pos2, pos3);
  return 0;
}