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 x[1000000];
int n;
bool f(long long d) {
  if (x[0] + 2 * d >= x[n - 1]) return true;
  int p1 = 0, p2 = n - 1;
  while (x[p1] <= x[0] + 2 * d) p1++;
  while (x[p2] >= x[n - 1] - 2 * d) p2--;
  return x[p2] - x[p1] <= 2 * d;
}
int main() {
  long long lo, hi, mid;
  cin >> n;
  for (int i = 0; i < (n); ++i) cin >> x[i], x[i] *= 2;
  sort(x, x + n);
  lo = 0, hi = INT_MAX;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    if (f(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  printf("%.6lf\n", lo / 2.);
  printf("%.6lf ", (x[0] + lo) / 2.);
  int p1 = 0, p2 = n - 1;
  while (x[p1] <= x[0] + 2 * lo) p1++;
  while (x[p2] >= x[n - 1] - 2 * lo) p2--;
  if (p1 > p2)
    printf("%.6lf", (x[0] + lo) / 2.);
  else
    printf("%.6lf", (x[p1] + x[p2]) / 4.);
  printf(" %.6lf\n", (x[n - 1] - lo) / 2.);
  return 0;
}