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, a[200005];
bool ok(int mid) {
  int temp = a[0] + mid, k = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] > temp) {
      if (k < 2) {
        temp = a[i] + mid;
        k++;
      } else
        return 0;
    }
  }
  return 1;
}
int main() {
  int st = 0, ed, mid, temp, k = 0;
  double answer;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  ed = a[n - 1];
  while (st < ed) {
    mid = (st + ed) / 2;
    if (ok(mid))
      ed = mid;
    else
      st = mid + 1;
  }
  temp = a[0] + st;
  answer = st / 2.0;
  cout << fixed << setprecision(6) << answer << endl;
  cout << fixed << setprecision(6) << a[0] + answer << " ";
  for (int i = 0; i < n; i++) {
    if (a[i] > temp) {
      temp = a[i] + st;
      k++;
      cout << fixed << setprecision(6) << a[i] + answer << " ";
    }
  }
  while (k < 2) {
    k++;
    cout << fixed << setprecision(6) << a[n - 1] / 1.0 << " ";
  }
  return 0;
}