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, arr[(int)2e5 + 5];
double l = 0, h = 10000000000, mid, ans;
bool GO(double m, int *arr, int no) {
  int x = arr[0];
  auto it = upper_bound(arr, arr + no, m * 2 + x);
  for (int i = 0; i < 2; ++i) {
    it = upper_bound(arr, arr + no, m * 2 + x);
    x = *it;
  }
  auto it1 = upper_bound(arr, arr + no, m * 2 + x);
  if (it < arr + no && ((m * 2) + *it) < arr[no - 1])
    return 0;
  else
    return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", arr + i);
  if (n == 3) {
    printf("%f\n", 0.0);
    printf("%.6f %.6f %.6f\n", (double)arr[0], (double)arr[1], (double)arr[2]);
    return 0;
  }
  sort(arr, arr + n);
  for (int i = 0; i < 1000; ++i) {
    mid = (l + h) / 2;
    if (GO(mid, arr, n)) {
      ans = mid;
      h = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  printf("%.6f\n", ans);
  auto it = upper_bound(arr, arr + n, arr[0] + ans * 2);
  auto it1 = upper_bound(arr, arr + n, *it + ans * 2);
  printf("%.6f %.6f %.6f\n", arr[0] + ans, *it + ans, *it1 + ans);
}