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;
int n;
int homes[200005];
bool can(long long d) {
  long long mxvis = 0, sta = 3;
  for (int i = 0; i < n; ++i) {
    if (homes[i] > mxvis) {
      if (sta < 1) return false;
      mxvis = homes[i] + 2 * d;
      --sta;
    }
  }
  return true;
}
long long bs() {
  long long lo = 0, hi = 2000000005, mid;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    if (can(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  return lo;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", homes + i), homes[i] *= 2;
  sort(homes, homes + n);
  long long d = bs();
  printf("%6f\n", (double)d / 2);
  int mxvis = 0, sta = 3;
  for (int i = 0; i < n; ++i) {
    if (homes[i] > mxvis && sta) {
      printf("%6f ", (double)(homes[i] + d) / 2), --sta,
          mxvis = homes[i] + 2 * d;
    }
  }
  while (sta) printf("%6f ", (double)homes[n - 1]), --sta;
  return 0;
}