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(void) {
  int n, i, a[200005];
  cin >> n;
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  if (n <= 3) {
    cout << 0 << endl;
    for (i = 0; i < 3; i++) printf("%.6lf ", double(a[min(i, n - 1)]));
    cout << endl;
    return 0;
  }
  double r = a[n - 1], l = 0, m;
  while (r - l > 1e-10) {
    m = (l + r) / 2;
    int sum = 1, zuo = 0;
    for (i = 0; i < n; i++)
      if (double(a[i] - a[zuo]) - 2 * m > 1e-9) {
        zuo = i;
        sum++;
      }
    if (sum > 3)
      l = m;
    else
      r = m;
  }
  int zuo = 0, sum = 0;
  printf("%.6lf\n", l);
  for (i = 0; i < n; i++)
    if (a[i] - a[zuo] - 2 * l > 1e-7) {
      printf("%.6lf ", double(a[i - 1] + a[zuo]) / 2);
      zuo = i;
      sum++;
    }
  printf("%.6lf", double(a[n - 1] + a[zuo]) / 2);
  sum++;
  if (sum < 3) {
    for (i = 0; i < 3 - sum; i++) printf(" %.6lf", double(a[n - 1]));
    cout << endl;
  } else
    printf("\n");
  return 0;
}