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;
const int N = 2e5 + 5;
const double eps = 1e-6;
const int mod = 1e9 + 7;
int n;
double a[N];
vector<double> qu;
bool check(double d) {
  qu.clear();
  qu.emplace_back(a[1] + d);
  int cnt = 1;
  for (int i = 1; i <= n; i++) {
    double dx = abs(a[i] - qu.back());
    if (dx - d >= eps) {
      cnt++;
      qu.emplace_back(a[i] + d);
    }
  }
  if (cnt <= 3)
    return true;
  else
    return false;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  double l = 0, r = 1e9, ans = 0;
  double a1 = 0, a2 = 0, a3 = 0;
  if (n == 3) {
    cout << "0" << endl;
    cout << a[1] << " " << a[2] << " " << a[3] << endl;
    return 0;
  }
  while (r - l > eps) {
    double mid = (l + r) / 2.0;
    if (check(mid)) {
      ans = mid;
      if (qu.size() == 1) {
        a1 = qu[0], a2 = a[2], a3 = a[3];
      } else if (qu.size() == 2) {
        a1 = qu[0], a2 = qu[1], a3 = a[n];
      } else {
        a1 = qu[0], a2 = qu[1], a3 = qu[2];
      }
      r = mid;
    } else
      l = mid;
  }
  printf("%.6f\n%.6f %.6f %.6f\n", ans, a1, a2, a3);
  return 0;
}