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;
const int M = 200000 + 10;
int a[M];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  double R_L = 0;
  double R_R = 0.5 * a[n - 1] - 0.5 * a[0];
  int num = 0;
  double C1, C2, C3;
  C2 = a[0];
  while (num < 200) {
    double R_M = (R_L + R_R) / 2;
    double D1 = a[0] + R_M * 2;
    int ind1 = -1;
    for (int i = 0; i < n; i++) {
      if (a[i] > D1) {
        ind1 = i;
        break;
      }
    }
    double D2 = R_M * 2;
    int ind2 = -1;
    for (int i = n - 1; i >= 0; i--) {
      if (a[i] + D2 < a[n - 1]) {
        ind2 = i;
        break;
      }
    }
    if (ind1 == -1 || ind2 == -1) {
      R_R = R_M;
    } else {
      if (ind1 > ind2) {
        R_R = R_M;
      } else {
        int Dis = a[ind2] - a[ind1];
        C2 = 0.5 * (a[ind2] + a[ind1]);
        if (Dis <= R_M * 2) {
          R_R = R_M;
        } else {
          R_L = R_M;
        }
      }
    }
    num++;
  }
  C1 = a[0] + R_R;
  C3 = a[n - 1] - R_R;
  C2 = a[0];
  for (int i = 0; i < n; i++) {
    if (a[i] > a[0] + 2 * R_R) {
      C2 = a[i] + R_R;
      break;
    }
  }
  printf("%.10f\n", R_R);
  printf("%.10f %.10f %.10f\n", C1, C2, C3);
  return 0;
}