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;
double a[300000];
int n;
int main() {
  int i, j, k;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%lf", a + i);
  sort(a, a + n);
  double low = 0, high = 1e10, mid;
  while (fabs(low - high) > 1e-9) {
    mid = (low + high) / 2.0;
    int tmp = 0;
    double pre = -1e10;
    for (i = 0; i < n; i++) {
      while (a[i] <= pre + 1e-8 && i < n) i++;
      if (i == n) break;
      tmp++;
      pre = a[i] + 2 * mid;
    }
    if (tmp <= 3) {
      high = mid;
    } else {
      low = mid;
    }
  }
  mid += 1e-8;
  printf("%lf\n", mid);
  double pre = -1e10;
  int tmp = 0;
  for (i = 0; i < n; i++) {
    while (a[i] <= pre + 1e-8 && i < n) i++;
    if (i == n) break;
    tmp++;
    pre = a[i] + 2 * mid;
    printf("%lf", pre - mid);
    if (tmp < 3)
      printf(" ");
    else {
      puts("");
      break;
    }
  }
  if (tmp == 2) {
    printf("%lf\n", 1.5e9);
  } else if (tmp == 1) {
    printf("%lf %lf\n", 1.5e9, 1.9e9);
  }
  return 0;
}