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 x[222222], n;
double d, x1, x2, x3, l, r, md;
bool check(double d) {
  bool f;
  int ps;
  x1 = x[1] + d;
  double xx = x1 + d;
  ps = 0, f = 0;
  for (int i = 1; i <= n; i++)
    if (xx - x[i] < -1e-9) {
      ps = i, f = 1;
      break;
    }
  if (!f) {
    x2 = x3 = x1;
    return 1;
  } else
    x2 = x[ps] + d, xx = x2 + d;
  f = 0;
  for (int i = ps; i <= n; i++)
    if (xx - x[i] < -1e-9) {
      ps = i, f = 1;
      break;
    }
  if (!f) {
    x3 = x2;
    return 1;
  } else
    x3 = x[ps] + d, xx = x3 + d;
  f = 0;
  for (int i = ps; i <= n; i++)
    if (xx - x[i] < -1e-9) {
      f = 1;
      break;
    }
  return !f;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  sort(x + 1, x + 1 + n);
  if (n == 1) {
    printf("0\n%.6lf %.6lf %.6lf\n", (double)x[1], (double)x[1], (double)x[1]);
    return 0;
  }
  if (n == 2) {
    printf("0\n%.6lf %.6lf %.6lf\n", (double)x[1], (double)x[1], (double)x[2]);
    return 0;
  }
  if (n == 3) {
    printf("0\n%.6lf %.6lf %.6lf\n", (double)x[1], (double)x[2], (double)x[3]);
    return 0;
  }
  l = 0, r = 2000000002.0;
  for (int ii = 0; ii < 1000; ii++) {
    md = (l + r) / 2.;
    if (check(md))
      r = md;
    else
      l = md;
  }
  if (!check(l)) l += 1e-6, check(l);
  printf("%.6lf\n%.6lf %.6lf %.6lf\n", l, x1, x2, x3);
  return 0;
}