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 N = 2e5 + 5;
int n, homes[N];
bool valid(double w) {
  double range = w;
  int testIndex = upper_bound(homes, homes + n, homes[0] + 2 * range) - homes;
  if (testIndex == n) return true;
  testIndex =
      upper_bound(homes, homes + n, homes[testIndex] + 2 * range) - homes;
  if (testIndex == n) return true;
  testIndex =
      upper_bound(homes, homes + n, homes[testIndex] + 2 * range) - homes;
  if (testIndex == n) return true;
  return false;
}
double minimumRange() {
  double low = 0.0, mid, high = 1000000005;
  int counter = 100;
  while (counter--) {
    mid = (low + high) / 2.0;
    if (valid(mid))
      high = mid;
    else
      low = mid;
  }
  return mid;
}
int main() {
  double EPS = 1e-6;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &homes[i]);
  sort(homes, homes + n);
  printf("%lf\n", minimumRange());
  printf("%lf ", homes[0] + minimumRange());
  int secondPlace =
      upper_bound(homes, homes + n, homes[0] + 2 * (minimumRange() + EPS)) -
      homes;
  printf("%lf ", homes[secondPlace] + minimumRange());
  int thirdPlace =
      upper_bound(homes, homes + n,
                  homes[secondPlace] + 2 * (minimumRange() + EPS)) -
      homes;
  printf("%lf\n", homes[thirdPlace] + minimumRange());
  return 0;
}