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 x[200005];
int res[3];
bool enough(long long d) {
  d *= 2;
  res[0] = x[0];
  int k = 0;
  for (int i = 0; i < n; ++i) {
    if (res[k] + d < x[i]) {
      ++k;
      if (k == 3) return false;
      res[k] = x[i];
    }
  }
  for (int i = k + 1; i < 3; ++i) res[i] = res[k];
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
  sort(x, x + n);
  for (int i = 0; i < n; ++i) x[i] *= 2;
  int l = 0, r = 2000000001, m;
  while (l < r) {
    m = ((long long)l + r) / 2;
    if (!enough(m))
      l = m + 1;
    else
      r = m;
  }
  enough(l);
  printf("%d.%d00000\n", l / 2, 5 * (l % 2));
  for (int i = 0; i < 3; ++i) {
    int x = res[i] / 2 + l / 2;
    printf("%d.%d00000%c", x, 5 * (l % 2), " \n"[i + 1 == 3]);
  }
  return 0;
}