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, t[210000], s[2 * 210000];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &t[i]);
  sort(t, t + n);
  for (int i = 0; i < n; i++) s[i] = t[i] - t[0], s[n + i] = t[n - 1] - t[i];
  sort(s, s + 2 * n);
  int i = 0, j = n - 1;
  double r = 1e20, x[3];
  for (int k = 0; k < 2 * n; k++) {
    int L = s[k];
    while (i < n && t[i] <= t[0] + L) i++;
    while (j >= 0 && t[n - 1] - L <= t[j]) j--;
    double rr = L;
    if (i <= j) rr = max(rr, (double)t[j] - t[i]);
    if (rr < r) {
      r = rr;
      x[0] = t[0] + r / 2;
      x[1] = t[n - 1] - r / 2;
      x[2] = i <= j ? 0.5 * (t[j] + t[i]) : t[0] + r / 2;
    }
  }
  printf("%.6lf\n%.6lf %.6lf %.6lf\n", r / 2, x[0], x[1], x[2]);
}