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;
int x[200000];
double eps = 1e-9;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> x[i];
  sort(x, x + n);
  double l = 0;
  double r = 2e9;
  int glFirst = -1;
  int first;
  int last;
  for (int it = 0; it < 200; it++) {
    double mid = (r + l) / 2;
    double x1 = x[0] + mid;
    double x2 = x[n - 1] - mid;
    first = -1;
    last = -1;
    for (int i = 0; i < n; i++) {
      if (fabs(x1 - x[i]) < mid || fabs(fabs(x1 - x[i]) - mid) < eps) continue;
      if (fabs(x2 - x[i]) < mid || fabs(fabs(x2 - x[i]) - mid) < eps) continue;
      if (first == -1) {
        first = i;
      }
      last = i;
    }
    if (first == -1) {
      r = mid;
    } else {
      double dist = x[last] - x[first];
      if (dist > 2 * mid && fabs(dist - 2 * mid) > eps) {
        l = mid;
      } else {
        r = mid;
        glFirst = first;
      }
    }
  }
  cout << fixed << setprecision(15) << l << "\n";
  cout << fixed << setprecision(15) << x[0] + l << " "
       << (glFirst == -1 ? x[0] + l : x[glFirst] + l) << " " << x[n - 1] - l;
  return 0;
}