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<double> v1;
double ans[3];
bool ok(double h) {
  int a1 = upper_bound(v1.begin(), v1.end(), v1[0] + 2 * h) - v1.begin();
  int a2 =
      lower_bound(v1.begin(), v1.end(), v1[v1.size() - 1] - 2 * h) - v1.begin();
  a2--;
  if (a2 <= a1) {
    ans[0] = v1[0] + h;
    ans[2] = v1[v1.size() - 1] - h;
    ans[1] = v1[a2];
    return true;
  } else {
    if (v1[a2] - v1[a1] <= 2 * h) {
      ans[0] = v1[0] + h;
      ans[2] = v1[v1.size() - 1] - h;
      ans[1] = v1[a1] + h;
      return true;
    } else
      return false;
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    double gg;
    scanf("%lf", &gg);
    v1.push_back(gg);
  }
  sort(v1.begin(), v1.end());
  double left = 0;
  double right = 4e9;
  double mid = (left + right) / 2;
  if (n == 1) {
    printf("0\n");
    printf("%.6lf %.6lf %.6lf\n", v1[0], v1[0], v1[0]);
    return 0;
  }
  if (n == 2) {
    printf("0\n");
    printf("%.6lf %.6lf %.6lf\n", v1[0], v1[0], v1[1]);
    return 0;
  }
  if (n == 3) {
    printf("0\n");
    printf("%.6lf %.6lf %.6lf\n", v1[0], v1[1], v1[2]);
    return 0;
  }
  for (int i = 0; i < 60; i++) {
    mid = (left + right) / 2;
    if (ok(mid)) {
      right = mid;
    } else
      left = mid;
  }
  printf("%.6lf\n", right);
  printf("%.6lf %.6lf %.6lf\n", ans[0], ans[1], ans[2]);
  return 0;
}