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[200001];
int n;
pair<pair<int, int>, bool> check(int d) {
  int sum = 0;
  int ct = 0;
  pair<pair<int, int>, bool> p = make_pair(make_pair(0, 0), 0);
  for (int i = 1; i < n; i++) {
    sum += a[i] - a[i - 1];
    if (sum > d) {
      ct++;
      sum = 0;
      if (ct == 1) {
        p.first.first = i - 1;
      } else if (ct == 2) {
        p.first.second = i - 1;
      } else {
        return p;
      }
    }
  }
  p.second = true;
  if (p.first.second == 0) {
    p.first.second = max(0, n - 2);
  }
  return p;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  a[n] = a[n - 1];
  int l = 0;
  int r = 2 * (a[n - 1] - a[0]);
  int m = (l + r) / 2;
  pair<pair<int, int>, bool> p;
  while (l < r) {
    p = check(m);
    if (p.second)
      r = m;
    else
      l = m + 1;
    m = (l + r) / 2;
  }
  p = check(m);
  cout << fixed;
  cout.precision(20);
  cout << double(m) / 2 << endl;
  cout << double(a[p.first.first] + a[0]) / 2 << " "
       << double(a[p.first.second] + a[p.first.first + 1]) / 2 << " "
       << double(a[n - 1] + a[p.first.second + 1]) / 2;
  return 0;
}