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 a[200001];
int main() {
  int n;
  while (scanf("%d", &n) >= 1) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    sort(a, a + n);
    double bp0 = a[0], bp1 = a[0], bp2 = a[0];
    int d = a[n - 1] - a[0];
    for (int i = 1; i < n; i++) {
      int l = i;
      int r = n - 1;
      while (r > l + 1) {
        int mid = (l + r) / 2;
        if ((a[mid] - a[i]) - (a[n - 1] - a[mid + 1]) > 0)
          r = mid;
        else
          l = mid;
      }
      int dn = max(a[i - 1] - a[0],
                   max(a[l] - a[i], a[n - 1] - a[min(l + 1, n - 1)]));
      if (d > dn) {
        d = dn;
        bp0 = (a[i - 1] + a[0] + 0.0) / 2;
        bp1 = (a[l] + a[i] + 0.0) / 2;
        bp2 = (a[n - 1] + a[min(l + 1, n - 1)] + 0.0) / 2;
      }
      l = i;
      r = n - 1;
      while (r > l + 1) {
        int mid = (l + r) / 2;
        if ((a[mid] - a[i]) - (a[n - 1] - a[mid + 1]) < 0)
          l = mid;
        else
          r = mid;
      }
      dn = max(a[i - 1] - a[0],
               max(a[r] - a[i], a[n - 1] - a[min(r + 1, n - 1)]));
      if (d > dn) {
        d = dn;
        bp0 = (a[i - 1] + a[0] + 0.0) / 2;
        bp1 = (a[r] + a[i] + 0.0) / 2;
        bp2 = (a[n - 1] + a[min(r + 1, n - 1)] + 0.0) / 2;
      }
    }
    if (n == 2) {
      bp2 = a[1];
      d = 0;
    }
    printf("%.6lf\n%.6lf %.6lf %.6lf\n", (d + 0.0) / 2, bp0, bp1, bp2);
  }
  return 0;
}