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;
int x[1000000];
int ini1, fin1, ini2, fin2, ini3, fin3;
int i1, i2;
bool sepuede(int d) {
  i1 = 0;
  while (i1 + 1 < n and x[i1 + 1] - x[0] <= d) i1++;
  ini1 = 0;
  fin1 = i1;
  if (i1 == n - 1) {
    ini2 = fin2 = ini3 = fin3 = n - 1;
    return true;
  }
  i2 = i1 + 1;
  while (i2 + 1 < n and x[i2 + 1] - x[i1 + 1] <= d) i2++;
  ini2 = i1 + 1;
  fin2 = i2;
  if (i2 == n - 1) {
    ini3 = fin3 = n - 1;
    return true;
  }
  if (x[n - 1] - x[i2 + 1] > d) return false;
  ini3 = i2 + 1;
  fin3 = n - 1;
  return true;
}
int main() {
  cout.setf(ios::fixed);
  cout.precision(10);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> x[i];
  sort(x, x + n);
  int inf = -1;
  int sup = 1000000000;
  while (inf + 1 < sup) {
    int med = (inf + sup) / 2;
    if (sepuede(med))
      sup = med;
    else
      inf = med;
  }
  sepuede(sup);
  double sol = max((x[fin1] - x[ini1]) / 2.0,
                   max((x[fin2] - x[ini2]) / 2.0, (x[fin3] - x[ini3]) / 2.0));
  cout << sol << endl;
  cout << (x[ini1] + x[fin1]) / 2.0 << " " << (x[ini2] + x[fin2]) / 2.0 << " "
       << (x[ini3] + x[fin3]) / 2.0 << endl;
}