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;
bool calc(const vector<double> &h, double d) {
  vector<double>::const_iterator it;
  it = upper_bound((h).begin(), (h).end(), h[0] + 2 * d);
  if (it == h.end()) return false;
  it = upper_bound((h).begin(), (h).end(), *it + 2 * d);
  if (it == h.end()) return false;
  it = upper_bound((h).begin(), (h).end(), *it + 2 * d);
  if (it == h.end()) return false;
  return true;
}
int main(void) {
  int n;
  double mn, mx;
  scanf("%d", &n);
  vector<double> h(n);
  for (int i = (0); i < ((n)); ++i) scanf("%lf", &h[i]);
  sort((h).begin(), (h).end());
  mn = 0;
  mx = h[n - 1];
  double md;
  for (int i = (0); i < ((100)); ++i) {
    md = (mx + mn) / 2;
    if (!calc(h, md))
      mx = md;
    else
      mn = md;
  }
  md = floor((md + 0.000001) * 2.0) / 2.0;
  printf("%.6f\n", md);
  printf("%.6f ", h[0] + md);
  vector<double>::iterator it =
      upper_bound((h).begin(), (h).end(), h[0] + 2 * md);
  if (it == h.end()) it = h.begin() + (n - 1);
  printf("%.6f ", *it + md);
  it = upper_bound((h).begin(), (h).end(), *it + 2 * md);
  if (it == h.end()) it = h.begin() + (n - 1);
  printf("%.6f", *it + md);
}