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>
#pragma comment(linker, "/STACK:20000000")
using namespace std;
long n, i;
double m, p1, p2, maxx, minn, r, l;
long ar[1000000];
int main() {
  cin >> n;
  cout.precision(6);
  for (i = 0; i < n; i++) {
    cin >> ar[i];
  }
  sort(ar, ar + n);
  l = 0;
  r = 2000000000;
  while (r - l > 0.01) {
    m = (l + r) / 2;
    p1 = ar[0] + 2 * m;
    p2 = ar[n - 1] - 2 * m;
    minn = 1100000000;
    maxx = -1100000000;
    for (i = 0; i < n; i++) {
      if (ar[i] > p1 && ar[i] < p2) {
        if (ar[i] < minn) minn = ar[i];
        if (ar[i] > maxx) maxx = ar[i];
      };
    }
    if (maxx - minn > 2100000000 || maxx - minn < 2 * m) {
      r = m;
    } else
      l = m;
    ;
  }
  long lq = m * 2 + 0.1;
  m = lq * 0.5;
  cout << fixed << m << endl << ar[0] + m;
  minn = 1100000000;
  for (i = 0; i < n; i++)
    if (ar[i] > ar[0] + 2 * m + 0.00000001 && ar[i] < minn) minn = ar[i];
  if (minn > 1000000000) minn = ar[0];
  cout << " " << minn + m << " " << ar[n - 1] - m << endl;
  cin.get();
  cin.get();
  cin.get();
  return 0;
}