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;
long long pos[100];
vector<long long> x(200005);
bool ok(long long p) {
  pos[0] = x[0] + p;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (x[i] > pos[cnt] + p) pos[++cnt] = x[i] + p;
  }
  return cnt <= 2;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    x[i] *= 1e6;
  }
  sort(x.begin(), x.begin() + n);
  long long lo = 0, hi = x[n - 1], mid;
  while (hi - lo > 1) {
    mid = (hi + lo) / 2;
    if (ok(mid))
      hi = mid;
    else
      lo = mid;
  }
  lo += 3;
  while (ok(lo) && lo) lo--;
  if (lo) lo++;
  ok(lo);
  printf("%.6f\n", (double)lo / 1e6);
  printf("%.6f %.6f %.6f", (double)pos[0] / 1e6, (double)pos[1] / 1e6,
         (double)pos[2] / 1e6);
  return 0;
}