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, pos[200001];
bool doable(int d) {
  int f = 0, s = 1;
  while (s < n && pos[s] - pos[f] <= 2 * d) ++s;
  if (s >= n) return true;
  f = s;
  s = f + 1;
  while (s < n && pos[s] - pos[f] <= 2 * d) ++s;
  if (s >= n) return true;
  f = s;
  s = f + 1;
  while (s < n && pos[s] - pos[f] <= 2 * d) ++s;
  if (s >= n) return true;
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &pos[i]);
    pos[i] *= 2;
  }
  sort(pos, pos + n);
  int left = 0, right = 2000000000;
  int can = -1;
  while (left <= right) {
    int dis = (left + right) / 2;
    if (doable(dis)) {
      right = dis - 1;
      can = dis;
    } else {
      left = dis + 1;
    }
  }
  double d = can * 0.5;
  printf("%.6f\n", d);
  for (int i = 0; i < n; ++i) pos[i] /= 2;
  int f = 0;
  int s = f + 1;
  while (s < n && pos[s] - pos[f] <= can) ++s;
  printf("%.6f ", pos[f] + d);
  f = s;
  s = f + 1;
  while (s < n && pos[s] - pos[f] <= can) ++s;
  if (f == n) f = n - 1;
  printf("%.6f ", pos[f] + d);
  f = s;
  s = f + 1;
  while (s < n && pos[s] - pos[f] <= can) ++s;
  if (f == n) f = n - 1;
  printf("%.6f\n", pos[f] + d);
  return 0;
}