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 n;
double dist(int x2, int x1) { return double(abs(x2 - x1)); }
int main() {
  scanf("%d", &n);
  vector<int> vx(n);
  for (int i = (0), _i = (n); i < _i; ++i) scanf("%d", &vx[i]);
  sort(vx.begin(), vx.end());
  if (n < 4) {
    vx.push_back(vx[n - 1]);
    vx.push_back(vx[n - 1]);
    printf("0.000000\n");
    printf("%.6llf ", double(vx[0]));
    printf("%.6llf ", double(vx[1]));
    printf("%.6llf\n", double(vx[2]));
    return 0;
  }
  int x = n - 3, y = n - 2, z = n - 1;
  int sx = x, sy = y;
  double sd = 1000000000;
  while (true) {
    double dx = dist(vx[0], vx[x]);
    double dy = dist(vx[x + 1], vx[y]);
    double dz = dist(vx[y + 1], vx[z]);
    if (dx > dy) {
      if (dx > dz) {
        if (dx < sd) {
          sd = dx;
          sx = x;
          sy = y;
        }
        x--;
      } else {
        if (dz < sd) {
          sd = dz;
          sx = x;
          sy = y;
        }
        break;
      }
    } else {
      if (dy > dz) {
        if (dy < sd) {
          sd = dy;
          sx = x;
          sy = y;
        }
        y--;
      } else {
        if (dz < sd) {
          sd = dz;
          sx = x;
          sy = y;
        }
        break;
      }
    }
  }
  printf("%.6llf\n", sd / 2.0);
  printf("%.6llf ", (double(vx[0] + vx[sx])) / 2.0);
  printf("%.6llf ", (double(vx[sx + 1] + vx[sy])) / 2.0);
  printf("%.6llf\n", (double(vx[sy + 1] + vx[z])) / 2.0);
}