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;
bool cmp(long long x, long long y) { return (x > y); }
int main() {
  long long n, k;
  cin >> n;
  long long a[n];
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  int l = 0;
  int h = a[n - 1] - a[0];
  while (l < h) {
    int m = (l + h) >> 1;
    int i, k;
    for (i = k = 0; k < 3; ++k) {
      int j;
      for (j = i + 1; j < n && a[j] - a[i] <= m; ++j)
        ;
      i = j;
    }
    if (i >= n)
      h = m;
    else
      l = m + 1;
  }
  cout << fixed << setprecision(6) << ((l)*0.5) << '\n';
  int j;
  for (int k = 0, i = 0; k < 3; ++k) {
    for (j = i + 1; j < n && a[j] - a[i] <= l; ++j)
      ;
    if (i == n)
      cout << 100000000 << " ";
    else {
      printf("%.6lf ", 0.5 * (a[j - 1] - a[i]) + a[i]);
      i = j;
    }
  }
  return 0;
}