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[200010];
int n;
double center[4];
int len_center;
bool solve(double d) {
  double le;
  int le_i = 0;
  int i;
  for (i = 1; i <= 3 && le_i < n; ++i) {
    le = a[le_i];
    center[i] = le + d;
    len_center = i;
    le += 2 * d + 1e-7;
    le_i = upper_bound(a, a + n, le) - a;
  }
  if (le_i < n) {
    return 0;
  } else {
    return 1;
  }
}
int main() {
  double L = 0, R;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  sort(a, a + n);
  n = unique(a, a + n) - a;
  R = a[n - 1] + 1;
  while (R - L > 1e-7) {
    double mid = (R + L) / 2;
    if (solve(mid)) {
      R = mid;
    } else {
      L = mid;
    }
  }
  printf("%f\n", R);
  solve(R);
  printf("%f", center[1]);
  for (int i = 2; i <= len_center; ++i) {
    printf(" %f", center[i]);
  }
  for (int i = len_center + 1; i <= 3; ++i) {
    printf(" %f", center[1]);
  }
  return 0;
}