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<int> v;
vector<double> ans, pos;
int n;
bool ok(double d) {
  pos.resize(3);
  int now = 0;
  for (int i = 0; i < 3; i++) {
    pos[i] = v[now] + d;
    vector<int>::iterator it;
    it = upper_bound(v.begin(), v.end(), v[now] + 2 * d);
    if (it == v.end()) {
      for (int j = i + 1; j < 3; j++) pos[j] = pos[i];
      return true;
    }
    now = it - v.begin();
  }
  return false;
}
int main() {
  scanf("%d", &n);
  v.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  n = v.size();
  vector<double> ans;
  if (n <= 3) {
    puts("0");
    for (int i = 0; i < n; i++) ans.push_back(v[i]);
    while (ans.size() < 3) ans.push_back(v.back());
    printf("%.6lf %.6lf %.6lf\n", ans[0], ans[1], ans[2]);
    return 0;
  }
  ans.resize(3);
  double left = 0, right = 1000000000, res = right;
  for (int times = 0; times < 50; times++) {
    double d = (left + right) / 2;
    if (ok(d)) {
      right = d;
      if (d < res) {
        res = d;
        ans = pos;
      }
    } else {
      left = d;
    }
  }
  printf("%.6lf\n%.6lf %.6lf %.6lf\n", res, ans[0], ans[1], ans[2]);
  return 0;
}