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 max3(int a, int b, int c) {
  int res = a;
  if (b > res) res = b;
  if (c > res) res = c;
  return res;
}
int main(void) {
  int x[200000 + 100];
  int n;
  scanf("%d", &n);
  for (register int i = (0); i < (int)n; ++i) scanf("%d", &x[i]);
  sort(x, x + n);
  if (n == 1) {
    printf("0.000000\n%.6lf %.6lf %.6lf", (double)x[0], (double)x[0],
           (double)x[0]);
    return 0;
  } else if (n == 2) {
    printf("0.000000\n%.6lf %.6lf %.6lf", (double)x[0], (double)x[1],
           (double)x[1]);
    return 0;
  } else if (n == 3) {
    printf("0.000000\n%.6lf %.6lf %.6lf", (double)x[0], (double)x[1],
           (double)x[2]);
    return 0;
  }
  double mind = (x[n - 1] - x[0]) * 2, bf = x[0], bs = x[0], bt = x[0];
  for (register int i = (0); i < (int)n - 2; ++i) {
    int mn = i + 1, mx = n - 1;
    while (mx - mn > 2) {
      int mid = (mx + mn) / 2;
      if (x[mid] - x[i + 1] > x[n - 1] - x[mid + 1])
        mx = mid;
      else
        mn = mid;
    }
    for (register int d = (0); d < (int)3; ++d) {
      if (mn >= n - 1) break;
      double tmp = max3(x[i] - x[0], x[mn] - x[i + 1], x[n - 1] - x[mn + 1]);
      if (tmp < mind) {
        mind = tmp;
        bf = (x[i] + x[0]) / 2.0;
        bs = (x[mn] + x[i + 1]) / 2.0;
        bt = (x[n - 1] + x[mn + 1]) / 2.0;
      }
      mn++;
    }
  }
  printf("%lf\n%.6lf %.6lf %.6lf", mind / 2.0, bf, bs, bt);
  return 0;
}