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;
double a[200100];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%lf", &a[i]);
  sort(a, a + n);
  double l = 0, r = a[n - 1] - a[0];
  double best = 1e40;
  double res1 = 0, res2 = 0, res3 = 0;
  for (int it = (0); it < (2000); ++it) {
    double m = (l + r) / 2.0;
    int left = upper_bound(a, a + n, a[0] + m + m) - a;
    if (left == n) {
      best = m;
      res1 = res2 = res3 = a[0] + m;
      r = m;
      continue;
    }
    int right = lower_bound(a, a + n, a[n - 1] - m - m) - a;
    right--;
    if (right < left) {
      best = m;
      res1 = res2 = a[0] + m;
      res3 = a[n - 1] - m;
      r = m;
      continue;
    }
    if (a[right] - a[left] <= 2 * m) {
      best = m;
      res1 = a[0] + m;
      res3 = a[n - 1] - m;
      res2 = a[left] + (a[right] - a[left]) / 2;
      r = m;
    } else
      l = m;
  }
  printf("%.6lf\n%.6lf %.6lf %.6lf\n", best, res1, res2, res3);
  return 0;
}