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;
vector<int> a(200000);
bool check(double c1, double n1) {
  int j = 0, f = 0, t = 0;
  while (j < n1 && f <= 3) {
    f++;
    double x = a[j] + c1 * 2;
    while (a[j] <= x && j < n1) {
      j++;
    }
    if (j == n1) t = 1;
  }
  return t == 1 && f <= 3;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int i;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (n <= 3) {
    cout << "0" << endl;
    if (n == 1) cout << a[0] << ' ' << a[0] << ' ' << a[0];
    if (n == 2) cout << a[0] << ' ' << a[1] << ' ' << a[1];
    if (n == 3) cout << a[0] << ' ' << a[1] << ' ' << a[2];
    return 0;
  }
  sort(&a[0], &a[n]);
  double l = 0, r = a[n - 1], ans = 0;
  while (r - l > 1e-7) {
    double c = (r + l) / 2;
    if (check(c, n)) {
      r = c;
      ans = c;
    } else
      l = c;
  }
  cout.precision(7);
  cout << fixed;
  cout << ans << endl;
  i = 0;
  int fl = 0;
  while (i < n) {
    fl++;
    cout << a[i] + ans << ' ';
    double l = a[i] + ans * 2;
    while (a[i] <= l && i < n) i++;
  }
  if (fl < 3) {
    for (i = fl; i < 3; i++) cout << a[n - 1] << ' ';
  }
  return 0;
}