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;
double a[200003];
double b[200003];
void mergesort(double *a, double *b, int n) {
  int i, i0, i1, iRight, iEnd, width, j;
  for (width = 1; width < n; width *= 2)
    for (i = 0; i + width < n; i += width * 2) {
      i0 = i;
      i1 = iRight = i + width;
      iEnd = min(i + width * 2, n);
      for (j = i; i0 < iRight; j++)
        if (i1 == iEnd || a[i0] < a[i1])
          b[j] = a[i0++];
        else
          b[j] = a[i1++];
      for (j = i; j < i1; j++) a[j] = b[j];
    }
}
long double pos1, pos2, pos3;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  int i;
  long double ep = 0.000000001;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    a[i] += ep;
  }
  mergesort(a, b, n);
  long double l, r, re;
  l = 0.0000000;
  r = 2000000000.0000;
  int it = 50;
  while (it > 0) {
    it--;
    re = (l + r) / 2.0;
    int pos = 0;
    long double p1 = -1.00;
    long double p2 = -1.00;
    long double p3 = -1.00;
    for (i = 1; i < n; i++) {
      if (((long double)a[i] - (long double)a[pos]) / 2.0 + ep > re) {
        if (p1 < -ep)
          p1 = (a[i - 1] + a[pos]) / 2.00;
        else if (p2 < -ep)
          p2 = (a[i - 1] + a[pos]) / 2.00;
        else
          break;
        pos = i;
      }
    }
    p3 = ((long double)a[n - 1] + (long double)a[pos]) / 2.00;
    if (i == n) {
      pos1 = p1;
      pos2 = p2;
      pos3 = p3;
      if (pos1 < -ep) pos1 = ep;
      if (pos2 < -ep) pos2 = ep;
      if (pos3 < -ep) pos3 = ep;
      r = re;
    } else
      l = re;
  }
  cout << setprecision(20) << r << endl;
  cout << setprecision(20) << pos1 << " " << pos2 << " " << pos3 << endl;
  return 0;
}