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 org[200010];
double f[200010][4];
int back[200010][4];
int n;
int query(int ind, int x, int y) {
  int g;
  int mid = (x + y) >> 1;
  int l = x, r = y;
  while (l < r) {
    mid = (l + r) >> 1;
    double t2 =
        max(f[((mid + 1 + r) >> 1)][ind],
            (org[y + 1] - org[min(((mid + 1 + r) >> 1) + 1, y + 1)]) / 2.0);
    double t1 = max(f[((mid + l) >> 1)][ind],
                    (org[y + 1] - org[min(((mid + l) >> 1) + 1, y + 1)]) / 2.0);
    double t3 =
        max(f[(mid)][ind], (org[y + 1] - org[min((mid) + 1, y + 1)]) / 2.0);
    if (t1 >= t3 && t2 >= t3) {
      if ((mid + 1 + r) >> 1 == r && (mid + l) >> 1 == l) break;
      r = (mid + 1 + r) >> 1;
      l = (mid + l) >> 1;
    } else if (t1 < t3)
      r = mid;
    else if (t3 > t2)
      l = mid + 1;
  }
  return mid;
}
void outit(int x, int y) {
  if (y != 1) {
    outit(back[x][y], y - 1);
    int tmp = min(x, (back[x][y] + 1));
    printf("%.6lf ", (org[x] - org[tmp]) / 2.0 + org[tmp]);
  } else
    printf("%.6lf ", (org[x] - org[1]) / 2.0 + org[1]);
}
void display() {
  for (int j = 1; j <= 3; j++) {
    for (int i = 1; i <= n; i++) printf("%.6lf ", f[i][j]);
    printf("\n");
  }
  for (int j = 1; j <= 3; j++) {
    for (int i = 1; i <= n; i++) printf("%d ", back[i][j]);
    printf("\n");
  }
}
int main(int argc, char *argv[]) {
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%lf", &org[i]);
    sort(org + 1, org + n + 1);
    int lne = 1;
    double a[200010];
    a[1] = org[1];
    for (int i = 2; i <= n; i++) {
      if (org[i] != org[i - 1]) a[++lne] = org[i];
    }
    for (int i = 1; i <= lne; i++) org[i] = a[i];
    n = lne;
    memset(f, 0, sizeof(f));
    memset(back, 0, sizeof(back));
    f[2][1] = (org[2] - org[1]) / 2.0;
    f[3][1] = (org[3] - org[1]) / 2.0;
    back[2][2] = 1;
    back[2][3] = 1;
    back[3][3] = 2;
    if ((org[2] - org[1]) / 2.0 < (org[3] - org[2]) / 2.0) {
      f[3][2] = (org[2] - org[1]) / 2.0;
      back[3][2] = 2;
    } else {
      f[3][2] = (org[3] - org[2]) / 2.0;
      back[3][2] = 1;
    }
    for (int i = 4; i <= n; i++) f[i][1] = (org[i] - org[1]) / 2.0;
    for (int j = 2; j <= 3; j++)
      for (int i = 4; i <= n; i++) {
        int z = query(j - 1, 1, i - 1);
        double t = max(f[z][j - 1], (org[i] - org[min(z + 1, i)]) / 2.0);
        f[i][j] = t;
        back[i][j] = z;
        if (f[i - 1][j - 1] < f[i][j]) {
          f[i][j] = f[i - 1][j - 1];
          back[i][j] = i - 1;
        }
      }
    printf("%.6lf\n", f[n][3]);
    outit(n, 3);
    printf("\n");
  }
  return 0;
}