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 n, i, j;
double ar[210000], ans;
vector<double> ans2;
double MIN(double a, double b) {
  if (a < b)
    return a;
  else
    return b;
}
double MAX(double a, double b) {
  if (a > b)
    return a;
  else
    return b;
}
bool cek(double v) {
  int ctr = 0;
  double k;
  bool f = 0;
  for (int i = 0; i < n; i = i) {
    k = ar[i] + v * 2;
    ctr++;
    while (ar[i] <= k && i < n) i++;
  }
  if (ctr <= 3)
    return true;
  else
    return false;
}
void print(double v) {
  int ctr = 0;
  double k;
  for (int i = 0; i < n; i = i) {
    k = ar[i] + v * 2;
    ans2.push_back(ar[i] + v);
    ctr++;
    while (ar[i] <= k && i < n) i++;
  }
  for (int i = 0; i < 3 - ctr; i++) ans2.push_back(ar[i]);
}
void bin_search(double l, double r) {
  if (fabs(l - r) <= 0.0000001) return;
  double p = (l + r) / 2;
  bool ck = cek(p);
  if (ck) {
    ans = p;
    bin_search(l, p);
  } else {
    bin_search(p, r);
  }
}
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%lf", &ar[i]);
  sort(ar, ar + n);
  bin_search(0, ar[n - 1] - ar[0]);
  printf("%.6lf\n", ans);
  cek(ans);
  print(ans);
  bool f = 0;
  sort(ans2.begin(), ans2.end());
  for (i = 0; i < 3; i++) {
    if (!f) {
      f = 1;
      printf("%.6lf", ans2[i]);
    } else
      printf(" %.6lf", ans2[i]);
  }
  cout << endl;
  return 0;
}