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 N1 = 205000;
long long point[N1];
double station[4];
bool seekOk(int m, long long d) {
  double r = double(d) / 2;
  int ith = 1;
  long long maxn = point[1] + d;
  station[1] = point[ith] + r;
  while (ith < m && point[ith] <= maxn) ith++;
  maxn = point[ith] + d;
  station[2] = point[ith] + r;
  while (ith < m && point[ith] <= maxn) ith++;
  maxn = point[ith] + d;
  station[3] = point[ith] + r;
  if (point[m] <= maxn)
    return true;
  else
    return false;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &point[i]);
  }
  sort(point + 1, point + 1 + n);
  int m = unique(point + 1, point + 1 + n) - 1 - point;
  long long lt = 0, rt = (point[m] - point[1]);
  while (rt > lt) {
    if (seekOk(m, ((lt + rt) >> 1)))
      rt = ((lt + rt) >> 1);
    else
      lt = ((lt + rt) >> 1) + 1;
  }
  if (!seekOk(m, lt)) seekOk(m, lt + 1);
  if (lt)
    printf("%.6f\n", double(lt) / 2);
  else
    printf("0\n");
  printf("%.6f %.6f %.6f\n", station[1], station[2], station[3]);
  return 0;
}