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;
int a[200005];
bool ok(double d) {
  int ct = 0;
  int i, j;
  for (i = 0; i < (n); i++) {
    for (j = i; j < n; j++) {
      if (a[j] - a[i] > 2 * d) break;
    }
    ct++;
    i = j - 1;
    if (ct > 3) return false;
  }
  return true;
}
int main() {
  double lo, hi, mid;
  int i, it, j;
  bool f;
  double d;
  int ct;
  while (scanf(" %d", &n) == 1) {
    for (i = 0; i < (n); i++) scanf(" %d", &a[i]);
    sort(a, a + n);
    lo = 0;
    hi = (a[n - 1] - a[0]) * .5;
    for (it = 0; it < 1000; it++) {
      if (fabs(lo - hi) < 1e-8) break;
      mid = (lo + hi) * .5;
      if (ok(mid))
        hi = mid;
      else
        lo = mid;
    }
    d = (lo + hi) * .5;
    printf("%.6lf\n", d);
    f = 0;
    ct = 0;
    for (i = 0; i < (n); i++) {
      for (j = i; j < n; j++) {
        if (a[j] - a[i] > 2 * d + 1E-7) break;
      }
      mid = (a[i] + a[j - 1]) * .5;
      if (!f)
        f = 1;
      else
        printf(" ");
      printf("%.6lf", mid);
      i = j - 1;
      ct++;
    }
    mid = (a[0] + a[n - 1]) * .5;
    for (i = 0; i < 3 - ct; i++) {
      if (!f)
        f = 1;
      else
        printf(" ");
      printf("%.6lf", mid);
    }
    printf("\n");
  }
  return 0;
}