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;
const int OO = (int)2e9;
const double eps = 1e-9;
double arr[200005];
int main() {
  int i, n, s, e, mid, startmid, endmid, curend;
  double mn, mx, end1, start3;
  scanf("%d", &n);
  for ((i) = 0; (i) < (n); (i)++) scanf("%lf", arr + i);
  sort(arr, arr + n);
  mx = arr[n - 1], mn = arr[0];
  s = 0, e = (mx - mn) * 2;
  while (s < e) {
    mid = s + (e - s) / 2;
    end1 = mn + mid / 2.0;
    start3 = mx - mid / 2.0;
    startmid = lower_bound(arr, arr + n, end1 + 0.5) - arr;
    endmid = lower_bound(arr, arr + n, start3) - arr - 1;
    curend = upper_bound(arr, arr + n, arr[startmid] + mid / 2.0) - arr - 1;
    if (curend >= endmid)
      e = mid;
    else
      s = mid + 1;
  }
  printf("%.6lf\n", s / 4.0);
  end1 = mn + s / 4.0;
  start3 = mx - s / 4.0;
  startmid = upper_bound(arr, arr + n, mn + s / 2.0) - arr;
  printf("%.6lf %.6lf %.6lf\n", end1, arr[startmid] + s / 4.0, start3);
  return 0;
}