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 main() {
  int n;
  double x[200001], eps = 1e-8;
  scanf("%d", &n);
  for (long long i = 0; i < (long long)n; i++) scanf("%lf", &x[i]);
  sort(x, x + n);
  double low = 0.0, high = 1e10;
  int i, j, k;
  while (true) {
    double mid = (low + high) / 2.0;
    if (low > high - eps) mid = high;
    for (i = 0; i < n; i++) {
      if (x[i] - x[0] > mid) break;
    }
    for (j = min(i, n - 1); j < n; j++) {
      if (x[j] - x[min(i, n - 1)] > mid) break;
    }
    for (k = min(j, n - 1); k < n; k++) {
      if (x[k] - x[min(j, n - 1)] > mid) break;
    }
    if (low > high - eps) break;
    if (k == n)
      high = mid;
    else
      low = mid;
  }
  printf("%.6f\n", high / 2.0);
  printf("%.6f %.6f %.6f\n", x[0] + (x[i - 1] - x[0]) / 2.0,
         x[min(i, n - 1)] + (x[j - 1] - x[min(i, n - 1)]) / 2.0,
         x[min(j, n - 1)] + (x[n - 1] - x[min(j, n - 1)]) / 2.0);
}