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 MAX = 1000000;
int MIN = -1000000;
int INF = 1000000000;
int x4[4] = {0, 1, 0, -1};
int y4[4] = {1, 0, -1, 0};
int x8[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int y8[8] = {1, 1, 0, -1, -1, -1, 0, 1};
int i, j, k;
double arr[500000], mid;
int n;
bool valid(double N) {
  int i;
  double papan1 = arr[0] + N;
  double papan2 = arr[n - 1] - N;
  double minim = 2000000000.0;
  double maksi = -1.0;
  for (i = (0); i < (n); i++) {
    if (arr[i] - papan1 > N && papan2 - arr[i] > N) {
      minim = (arr[i] < minim) ? arr[i] : minim;
      maksi = (arr[i] > maksi) ? arr[i] : maksi;
    }
  }
  if (maksi < 0) return true;
  if (maksi - minim > (N * 2.0)) return false;
  return true;
}
int main() {
  int j;
  cin >> n;
  for (i = (0); i < (n); i++) scanf("%lf", &arr[i]);
  sort(arr, arr + n);
  if (n <= 3) {
    cout << "0.000000" << endl;
    if (n == 3) printf("%.6lf %.6lf %.6lf\n", arr[0], arr[1], arr[2]);
    if (n == 2) printf("%.6lf %.6lf %.6lf\n", arr[0], arr[1], arr[1]);
    if (n == 1) printf("%.6lf %.6lf %.6lf\n", arr[0], arr[0], arr[0]);
    {
      fflush(stdin);
      getchar();
    };
    return 0;
  }
  double L = 0.0, R = 1000000000.0, ans = 1000000000.0;
  for (i = (0); i < (50); i++) {
    double mid = (L + R) / (double)2.0;
    if (valid(mid) == true) {
      R = mid;
      if (ans - mid >= 0) ans = mid;
    } else
      L = mid;
  }
  printf("%.6lf\n", ans);
  double papan1 = arr[0] + ans;
  double papan2 = arr[n - 1] - ans;
  double minim = 2000000000.0;
  double maksi = -1.0;
  for (i = (0); i < (n); i++) {
    if (arr[i] - papan1 > ans && papan2 - arr[i] > ans) {
      minim = (arr[i] < minim) ? arr[i] : minim;
      maksi = (arr[i] > maksi) ? arr[i] : maksi;
    }
  }
  if (maksi >= 0)
    printf("%.6lf %.6lf %.6lf\n", papan1, (maksi + minim) / 2.0, papan2);
  else
    printf("%.6lf %.6lf %.6lf\n", papan1, papan1, papan2);
  {
    fflush(stdin);
    getchar();
  };
  return 0;
}