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 x[200005];
int main() {
  memset(x, 0, sizeof(x));
  int n = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x[i]);
  }
  sort(x, x + n);
  double eps = 1e-10, l = 0.0, r = (double)(x[n - 1] - x[0]) / 6.0;
  int p1 = 0, p2 = 0;
  while (r - l > eps) {
    double mid = l + (r - l) / 2.0;
    p1 = upper_bound(x, x + n, x[0] + 2 * mid) - x;
    if (p1 == n) {
      r = mid;
      continue;
    }
    p2 = upper_bound(x, x + n, x[p1] + 2 * mid) - x;
    if (p2 == n || (double)(x[n - 1] - x[p2]) <= 2 * mid - eps) {
      r = mid;
    } else {
      l = mid;
    }
  }
  double tmp = x[0] + 2 * l;
  int idx = lower_bound(x, x + n, (int)(tmp - 1)) - x;
  for (; idx < n; ++idx) {
    if (x[idx] > x[0] + (int)(2 * l + 0.1)) {
      break;
    }
  }
  printf("%.6lf\n", l);
  printf("%.6lf %.6lf %.6lf", (double)x[0] + l, (double)x[idx] + l,
         (double)x[n - 1] - l);
  return 0;
}