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 a[200005], n;
double ans[3];
const double eps = 1e-10;
bool check(double r) {
  ans[0] = a[0] + r;
  ans[2] = a[n - 1] - r;
  int lo = 0, hi = n, L, R;
  while (lo + 1 < hi) {
    int m = (lo + hi) / 2;
    if (a[m] < ans[0] + r + eps)
      lo = m;
    else
      hi = m;
  }
  L = lo + 1;
  lo = -1, hi = n - 1;
  while (lo + 1 < hi) {
    int m = (lo + hi) / 2;
    if (a[m] > ans[2] - r - eps)
      hi = m;
    else
      lo = m;
  }
  R = hi - 1;
  if (L > R) {
    ans[1] = a[0];
    return 1;
  }
  ans[1] = (a[R] + a[L]) / 2.0;
  if (a[R] - a[L] > 2 * r + eps) return 0;
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  if (n <= 3) {
    printf("%.10f\n", 0.0);
    for (int i = 0; i < 3; i++)
      printf("%.10f%c", 1.0 * a[min(i, n - 1)], i == 2 ? '\n' : ' ');
    return 0;
  }
  double lo = 0, hi = (a[n - 1] - a[0]) / 2.0;
  for (int i = 0; i < 200; i++) {
    double m = (lo + hi) / 2;
    if (check(m))
      hi = m;
    else
      lo = m;
  }
  check(hi);
  printf("%.10f\n", hi);
  for (int i = 0; i < 3; i++) printf("%.10f%c", ans[i], i == 2 ? '\n' : ' ');
  return 0;
}