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 maxN = 2e5 + 10;
int n, a[maxN];
long double res[3];
bool check(long double x) {
  int i;
  for (i = 0; i < n; i++)
    if (a[i] > 2 * x + a[0]) break;
  res[0] = a[0] + x;
  int last = i;
  for (i; i < n; i++)
    if (a[i] > 2 * x + a[last]) break;
  res[1] = a[last] + x;
  last = i;
  for (i; i < n; i++)
    if (a[i] > 2 * x + a[last]) break;
  res[2] = a[last] + x;
  return (i >= n);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  if (n < 4)
    return cout << 0 << '\n' << a[0] << " " << a[1] << " " << a[2] << '\n', 0;
  long double l = 0, r = 1e14;
  for (int i = 0; i < 200; i++) {
    long double mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  cout << fixed << setprecision(10);
  if (check(double(r - 0.5)))
    cout << (double)r - 0.5 << '\n'
         << res[0] << " " << res[1] << " " << res[2] << '\n';
  else {
    check(r);
    cout << r << '\n' << res[0] << " " << res[1] << " " << res[2] << '\n';
  }
  return 0;
}