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<double> x;
int n;
bool good(double s) {
  int t = 0, c = 0;
  while (t < n && c < 3) {
    t = upper_bound(x.begin(), x.end(), x[t] + 2 * s + 1e-7) - x.begin();
    c++;
  }
  return t >= n;
}
int main() {
  scanf("%d", &n);
  x.resize(n + 1);
  for (int i = 0, _n = (n); i < _n; i++) scanf("%lf", &x[i]);
  x[n] = 1e11;
  sort(x.begin(), x.end());
  double l = 0, h = 1e9;
  while (h - l > 1e-7) {
    double m = (l + h) / 2.0;
    if (good(m))
      h = m;
    else
      l = m;
  }
  printf("%lf\n", l);
  int t = 0, c = 0;
  while (t < n && c < 3) {
    printf("%.6lf ", x[t] + l);
    t = upper_bound(x.begin(), x.end(), x[t] + 2 * l + 1e-6) - x.begin();
    c++;
  }
  if (c < 3)
    for (int i = (c), _b = (3); i < _b; i++) printf("%.6lf ", x[n - 1]);
  return 0;
}