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;
vector<long double> arr;
int n, ans;
long double l = 0, r = 999999999999;
long double lm = 0;
int x, y;
bool valid(long double d) {
  if (arr[arr.size() - 1] - 2 * d <= arr[0]) return true;
  x = *(lower_bound(arr.begin(), arr.end(), arr[arr.size() - 1] - 2 * d) - 1);
  y = *upper_bound(arr.begin(), arr.end(), arr[0] + 2 * d);
  if ((x - y) <= 2 * d) {
    return true;
  } else
    return false;
}
int main() {
  cin >> n;
  for (size_t i = 0; i < n; i++) {
    cin >> x;
    arr.push_back(x);
  }
  sort(arr.begin(), arr.end());
  r = arr[n - 1];
  l = 0;
  while (true) {
    long double mid = (l + r) / 2;
    if (abs(l - r) < 0.0000001) break;
    if (valid(mid)) {
      r = mid;
    } else {
      l = mid;
    }
    lm = r;
  }
  cout << fixed << setprecision(6) << (lm) << endl;
  cout << fixed << setprecision(6) << (long double)arr[0] + lm << " ";
  long double xx = (long double)arr[0] + lm;
  int i = 0, cnt = 1;
  for (size_t i = 0; i < n; i++) {
    if (arr[i] > xx + lm) {
      xx = arr[i] + lm;
      cout << fixed << setprecision(6) << (long double)arr[i] + lm << " ";
      cnt++;
    }
  }
  for (; cnt < 3; cnt++) cout << "0 ";
  cout << endl;
  return 0;
}