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 main() {
  vector<long long> a;
  int n;
  scanf("%d", &n);
  int x;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    a.push_back((long long)x * 2);
  }
  sort(a.begin(), a.end());
  long long left = -1;
  long long right = 2000000000;
  right *= 6;
  while (left + 1 < right) {
    long long mid = (left + right) / 2;
    int pos = 0;
    for (int vvv = 0; vvv < 3; vvv++) {
      int i;
      for (i = pos; i < n && a[pos] + 2 * mid >= a[i]; i++) {
      }
      pos = i;
    }
    if (pos == n)
      right = mid;
    else
      left = mid;
  }
  double ans = right;
  cout.precision(10);
  cout << ans / 2 << endl;
  int pos = 0;
  for (int vvv = 0; vvv < 3; vvv++) {
    if (pos == n)
      cout << a[n - 1] / 2 << " ";
    else
      cout << (ans + a[pos]) / 2 << " ";
    int i;
    for (i = pos; i < n && a[pos] + 2 * right >= a[i]; i++) {
    }
    pos = i;
  }
  return 0;
}