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 INF = 2147483647;
const double PI = 3.141592653589793;
const int N = 200005;
const double eps = 1e-7;
int x[N], tab[N], i, d, n;
double r, res[5];
void ustaw(double ind) {
  int maxx, i;
  res[0] = min(2000000000.0, tab[0] + ind);
  res[2] = max(0.0, tab[n - 1] - ind);
  maxx = 0;
  for (i = 0; i < n; i++)
    if (tab[i] > res[0] + ind + eps && tab[i] < res[2] - ind - eps)
      maxx = max(maxx, tab[i]);
  res[1] = max(0.0, maxx - ind);
}
bool bad(int a, double ind) {
  for (int i = 0; i < 3; i++)
    if (tab[a] < res[i] + ind + eps && tab[a] > res[i] - ind - eps)
      return false;
  return true;
}
double jedz(double a, double b) {
  double ind = (a + b) / 2;
  if (b - a < eps) return ind;
  int i, da;
  res[0] = tab[0] + ind;
  res[2] = tab[n - 1] - ind;
  ustaw(ind);
  da = 1;
  for (i = 0; i < n; i++)
    if (bad(i, ind)) da = 0;
  if (!da) return jedz(ind, b);
  return jedz(a, ind);
}
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &x[i]);
  sort(x, x + n);
  d = 1;
  tab[0] = x[0];
  for (i = 1; i < n; i++)
    if (x[i] != x[i - 1]) tab[d++] = x[i];
  n = d;
  if (n <= 3) {
    r = 0.0;
    if (n == 1) {
      res[0] = res[1] = res[2] = tab[0];
    }
    if (n == 2) {
      res[0] = res[1] = tab[0];
      res[2] = tab[1];
    }
    if (n == 3) {
      res[0] = tab[0];
      res[1] = tab[1];
      res[2] = tab[2];
    }
  } else
    r = jedz(0.0, INF / 2);
  printf("%.6lf\n", r);
  ustaw(r);
  sort(res, res + 3);
  for (i = 0; i < 3; i++) printf("%.6lf ", res[i]);
  printf("\n");
  return 0;
}