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;
long long int a[200001], n;
double temp[10], ans[10];
int ok(double d) {
  long long int cur = 0, i;
  for (i = 0; i < 3; i++) {
    temp[i] = a[cur] + d;
    cur = lower_bound(a, a + n, a[cur] + 2 * d + 1e-9) - a;
    if (cur >= n) break;
  }
  if (cur >= n)
    for (i = 0; i < 3; i++) ans[i] = temp[i];
  return cur >= n;
}
int main() {
  cout << fixed << setprecision(6);
  long long int i, j;
  double lo, hi, mid;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  lo = 0;
  hi = a[n - 1];
  for (i = 0; i < 200; i++) {
    mid = (lo + hi) / 2;
    if (ok(mid))
      hi = mid;
    else
      lo = mid;
  }
  cout << hi << endl;
  for (i = 0; i < 3; i++) cout << ans[i] << " ";
  return 0;
}