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 main() {
  int n, p[200000];
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); i++) scanf("%d", p + i);
  sort(p, p + n);
  double a = 0.0, b = (double)(p[n - 1] - p[0]) / 3.0 + 1;
  double t[3], m;
  for (int i = 0; i < (int)(3); i++) t[i] = 0.0;
  while (b - a > 1e-10) {
    m = (a + b) / 2.0;
    int base = 0;
    t[0] = p[0] + m;
    for (int i = 0; i < (int)(n); i++) {
      if ((double)p[i] - 1e-10 >= t[base] + m) {
        if (base == 2) {
          a = m;
          goto NEXT;
        }
        t[++base] = p[i] + m;
      }
    }
    b = m;
  NEXT:;
  }
  int base = 0;
  t[0] = p[0] + m;
  for (int i = 0; i < (int)(n); i++) {
    if (p[i] > t[base] + m + 1e-7) t[++base] = p[i] + m;
  }
  printf("%.6lf\n%.6lf %.6lf %.6lf\n", m, t[0], t[1], t[2]);
  return 0;
}