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;
double l, r, d;
vector<double> x, k, res;
int n;
int main() {
  cin >> n;
  k.resize(n);
  for (int i = 0; i < n; i++) cin >> k[i];
  sort(k.begin(), k.end());
  x.push_back(k[0]);
  for (int i = 1; i < k.size(); i++)
    if (k[i] != k[i - 1]) x.push_back(k[i]);
  double l = 0, r = x[x.size() - 1] - x[0], m;
  while (r - l >= 0.000000001) {
    m = (l + r) / 2.0;
    res.clear();
    for (int i = 0; i < x.size(); i++) {
      res.push_back(x[i] + m);
      int j = i;
      while (j < x.size() && x[j] - res[res.size() - 1] <= m) j++;
      i = j - 1;
    }
    if (res.size() <= 3)
      r = m + 0.0000000001;
    else
      l = m;
  }
  m = r;
  res.clear();
  for (int i = 0; i < x.size(); i++) {
    res.push_back(x[i] + m);
    int j = i;
    while (j < x.size() && x[j] - res[res.size() - 1] - m <= 0) j++;
    i = j - 1;
  }
  while (res.size() < 3) res.push_back(1);
  printf("%.9f\n%.9f %.9f %.9f", m, res[0], res[1], res[2]);
}