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;
const double eps = 1e-6;
int sh[200005], n;
double pos[5];
bool valid(double d) {
  pos[0] = sh[0] + d;
  int count = 1;
  double y = pos[0] + d;
  for (int i = 1; i < n && count <= 3; i++) {
    if (sh[i] - y < eps)
      continue;
    else {
      y = sh[i] + 2 * d;
      pos[count] = sh[i] + d;
      count++;
    }
  }
  return count <= 3;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> sh[i];
  sort(sh, sh + n);
  double start = 0, end = sh[0] + sh[n - 1], ans = 0;
  while (fabs(end - start) > eps) {
    double mid = (start + end) / 2;
    if (valid(mid)) {
      end = mid;
      ans = mid;
    } else {
      start = mid;
    }
  }
  cout << fixed << setprecision(10) << ans << endl;
  valid(ans);
  for (int i = 0; i < 3; i++) cout << pos[i] << " \n"[i == 2];
}