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;
double a[200001];
double ff[4];
int bins(double key, int n) {
  int l = 0;
  int r = n - 1;
  while (l <= r) {
    int c = l + (r - l) / 2;
    if (a[c] > key)
      r = c - 1;
    else
      l = c + 1;
  }
  return l;
}
bool check(double eps, int n) {
  int t;
  double p = a[0];
  ff[0] = a[0] + eps;
  t = bins(2 * eps + p, n);
  if (t > n - 1) return false;
  ff[1] = a[t] + eps;
  p = a[t] + 2 * eps;
  t = bins(p, n);
  if (t > n - 1) return false;
  ff[2] = a[t] + eps;
  p = a[t] + 2 * eps;
  t = bins(p, n);
  if (t > n - 1) return false;
  return true;
}
double ans(int n) {
  double l = 0;
  double r = a[n - 1];
  for (int i = 0; i < 70; i++) {
    double c = (l + r) / 2;
    if (!check(c, n))
      r = c;
    else
      l = c;
  }
  return l;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%lf", a + i);
  sort(a, a + n);
  if (n < 4) {
    printf("0.000000\n");
    for (int i = 0; i < 3 - n; i++)
      cout << "0.000000"
           << " ";
    for (int i = 0; i < n; i++) printf("%.6lf ", a[i]);
  } else {
    double q = ans(n);
    printf("%.6lf\n", q);
    for (int i = 0; i < 3; i++) printf("%.6lf ", ff[i]);
  }
}