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;
const int N = int(2e5) + 10;
int x[N], n;
bool check(int d) {
  int cnt = 0;
  int l = 0, r = 0;
  while (l < n) {
    while (r < n && x[r] - x[l] <= d) {
      r++;
    }
    l = r;
    cnt++;
  }
  return cnt <= 3;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  std::cout << std::fixed;
  std::cout << std::setprecision(6);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
  }
  sort(x, x + n);
  int high = int(1e9) + 5, low = 0, ans = -1;
  while (low <= high) {
    int mid = low + (high - low) / 2;
    if (check(mid)) {
      ans = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  double temp = ans / (double)2;
  cout << temp << "\n";
  int l = 0, r = 0;
  int val = 0;
  double last = 0;
  while (l < n) {
    last = double(x[l] + temp);
    cout << double(x[l] + temp) << " ";
    val++;
    while (r < n && x[r] - x[l] <= ans) {
      r++;
    }
    l = r;
  }
  while (val < 3) {
    cout << last << " ";
    val++;
  }
  return 0;
}