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, a[200009];
bool double_compare(int a, double b) {
  double tmp = (b + (1e-6)) * 1000000.0;
  long long tmp2 = tmp, tmp3 = (long long)a * 1000000;
  return tmp3 > tmp2;
}
bool valid(double d) {
  double pos = (double)(a[0]) + d;
  int num = 1;
  for (int i = 1; i < n; ++i) {
    double tmp = (pos + d) * 1000000.0;
    long long tmp2 = tmp, tmp3 = (long long)a[i] * 1000000;
    if (tmp3 > tmp2) {
      ++num;
      pos = (double)(a[i]) + d;
    }
  }
  return num <= 3;
}
void print_pos(double d) {
  double pos = (double)(a[0]) + d;
  printf("%.6lf", pos);
  int num = 1;
  for (int i = 1; i < n; ++i) {
    if (double_compare(a[i], pos + d)) {
      ++num;
      pos = (double)(a[i]) + d;
      printf(" %.6lf", pos);
    }
  }
  while (num < 3) ++num, printf(" %.6lf", pos);
  printf("\n");
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  double s = 0, e = 2e10;
  for (double mid = (e - s) / 2; mid >= 1e-8; mid *= 0.5) {
    if (!valid(s + mid)) s += mid;
  }
  printf("%.6lf\n", s);
  print_pos(s);
  return 0;
}