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 n, arr[200010];
double sol[3], tmp[3];
bool v(double d) {
  double end = 0;
  int j = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] > end) {
      if (j == 3) return 0;
      tmp[j++] = d + arr[i];
      end = arr[i] + 2.0 * d;
    }
  }
  for (int i = 0; i < 3; i++) sol[i] = tmp[i];
  return 1;
}
double S() {
  double l = 0, r = arr[n - 1];
  for (int i = 0; i < 400; i++) {
    double mid = (l + r) / 2;
    if (v(mid))
      r = mid;
    else
      l = mid;
  }
  return l;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  printf("%.6lf\n", S());
  printf("%.6lf %.6lf %.6lf", sol[0], sol[1], sol[2]);
  return 0;
}