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 loc[200005], n;
bool check(double x) {
  double T = 2 * x + loc[0];
  int ret = 1;
  for (int i = 0; i < n; i++) {
    if (loc[i] > T) {
      T = 2 * x + loc[i];
      ret++;
    }
  }
  return ret <= 3;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> loc[i];
  sort(loc, loc + n);
  double s = 0, e = 1e18, mid;
  double d = -1;
  while (s <= e) {
    mid = (s + e) / 2;
    if (check(mid))
      e = mid - 0.0000001, d = mid;
    else
      s = mid + 0.0000001;
  }
  cout << fixed << setprecision(6) << d << endl;
  double T = 2 * d + loc[0];
  cout << fixed << setprecision(6) << d + loc[0] << " ";
  int cnt = 1;
  for (int i = 0; i < n; ++i) {
    if (loc[i] > T) {
      T = 2 * d + loc[i];
      cout << fixed << setprecision(6) << d + loc[i] << " ";
      cnt++;
    }
  }
  if (cnt < 3) {
    while (cnt != 3) {
      cnt++;
      cout << fixed << setprecision(6) << (double)loc[n - 1] << " ";
    }
  }
  cout << endl;
}