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[2 * 100005];
double B[5];
bool ok(double x) {
  int counter = 2;
  double v = A[0] + (2 * x);
  for (int i = 1; i < n; ++i) {
    if (A[i] > v) {
      --counter;
      v = A[i] + (2 * x);
    }
  }
  if (counter < 0) {
    return false;
  } else {
    return true;
  }
}
double binarySearch() {
  double lo = 0, med, hi = 1e9;
  int cnt = 100;
  while (cnt--) {
    med = ((lo + hi) / 2);
    if (ok(med))
      hi = med;
    else
      lo = med;
  }
  return med;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
  }
  sort(A, A + n);
  double res = binarySearch();
  cout << fixed << setprecision(6) << res << endl;
  B[0] = A[0] + res;
  for (int i = 1, j = 0; i < n; ++i) {
    if (A[i] > (B[j] + res)) {
      B[j + 1] = A[i] + res;
      ++j;
    }
  }
  cout << fixed << setprecision(6) << B[0] << " " << B[1] << " " << B[2]
       << endl;
  return 0;
}