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 const inf = 1000000000;
vector<int> V;
bool ok(int mid) {
  int c = V[0], i;
  for (i = 0; i < V.size(); i++)
    if (V[i] - c > mid) break;
  if (i < V.size()) {
    c = V[i];
    for (; i < V.size(); i++)
      if (V[i] - c > mid) break;
    if (i < V.size()) {
      c = V[i];
      for (; i < V.size(); i++)
        if (V[i] - c > mid) break;
    }
  }
  if (i == V.size()) return true;
  return false;
}
int main() {
  V.clear();
  int n;
  cin >> n;
  int i;
  for (i = 0; i < n; i++) {
    int a;
    cin >> a;
    V.push_back(a);
  }
  sort(V.begin(), V.end());
  int low = 0;
  int high = 2000000000;
  int t = 100;
  while (t--) {
    int mid = (low + high) / 2;
    if (ok(mid))
      high = mid;
    else
      low = mid;
  }
  int k = max(0, low - 5);
  for (i = k;; i++) {
    if (ok(i)) break;
  }
  int mid, c;
  mid = i;
  printf("%.6lf\n", i / 2.0);
  c = V[0];
  for (i = 0; i < V.size(); i++)
    if (V[i] - c > mid) break;
  printf("%.6lf", (V[i - 1] - c) / 2.0 + c);
  if (i < V.size()) {
    c = V[i];
    for (; i < V.size(); i++)
      if (V[i] - c > mid) break;
    printf(" %.6lf", (V[i - 1] - c) / 2.0 + c);
    if (i < V.size()) {
      c = V[i];
      for (; i < V.size(); i++)
        if (V[i] - c > mid) break;
      assert(i == V.size());
      printf(" %.6lf", (V[i - 1] - c) / 2.0 + c);
    } else {
      printf(" %.6lf", (V[i - 1] - c) / 2.0 + c);
    }
  } else {
    printf(" %.6lf %.6lf\n", (V[i - 1] - c) / 2.0 + c,
           (V[i - 1] - c) / 2.0 + c);
  }
  return 0;
}