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;
long long max(long long a, long long b) {
  if (a < b) return b;
  return a;
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
bool okay(long long arr[], long long n, double d) {
  if (n <= 3) return true;
  double p = 2.0 * d + arr[0];
  long long i = 0;
  while (i < n && (double)arr[i] <= p) i++;
  if (i == n) return true;
  p = 2.0 * d + arr[i];
  while (i < n && (double)arr[i] <= p) i++;
  if (i == n) return true;
  p = 2.0 * d + arr[i];
  while (i < n && (double)arr[i] <= p) i++;
  return i == n;
}
void solve() {
  long long n;
  cin >> n;
  long long arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  if (n <= 3) {
    cout << 0 << "\n";
    if (n == 1)
      cout << arr[0] << " " << arr[0] << " " << arr[0] << "\n";
    else if (n == 2)
      cout << arr[0] << " " << arr[1] << " " << arr[1] << "\n";
    else
      cout << arr[0] << " " << arr[1] << " " << arr[2] << "\n";
    return;
  }
  double ans = 1e9;
  double low = 0.0, high = 1000000000.0;
  while (high - low > 1e-7) {
    double mid = low * 1.0 + 0.5 * (high - low);
    if (okay(arr, n, mid)) {
      high = mid;
    } else
      low = mid;
  }
  printf("%0.7f\n", high);
  double d = high;
  double p = 2.0 * d + arr[0];
  long long i = 0;
  printf("%0.7f ", d + arr[min(i, n - 1)]);
  while (i < n && (double)arr[i] <= p) i++;
  p = 2.0 * d + arr[min(i, n - 1)];
  printf("%0.7f ", d + arr[min(i, n - 1)]);
  while (i < n && (double)arr[i] <= p) i++;
  p = 2.0 * d + arr[min(i, n - 1)];
  printf("%0.7f ", d + arr[min(i, n - 1)]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}