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 main() {
  int n;
  scanf("%d", &n);
  vector<int> x;
  for (int i = 0; i < n; ++i) {
    int aa;
    scanf("%d", &aa);
    x.push_back(aa);
  }
  sort(x.begin(), x.end());
  int a = 0;
  int b = n - 1;
  int d = 0;
  int rd = INT_MAX;
  int ra = -1;
  int rb = -1;
  while (a + 1 < b && x[b - 1] - x[a + 1] > d) {
    if (x[b - 1] - x[a + 1] < rd) {
      rd = x[b - 1] - x[a + 1];
      ra = a;
      rb = b;
    }
    int d1 = x[a + 1] - x[0];
    int d2 = x[n - 1] - x[b - 1];
    if (d1 < d2) {
      d = max(d, d1);
      ++a;
    } else {
      d = max(d, d2);
      --b;
    }
  }
  if (d < rd) {
    rd = d;
    ra = a;
    rb = b;
  }
  printf("%.6f\n", rd / 2.0);
  printf("%.6f %.6f %.6f\n", (x[ra] + x[0]) / 2.0, (x[rb] + x[n - 1]) / 2.0,
         (ra + 1 < rb ? (x[ra + 1] + x[rb - 1]) / 2.0 : 0.0));
}