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 double eps = 1e-7;
const int M = 2 * 100000 + 2;
int dis[M];
int n;
void work() {
  sort(dis, dis + n);
  int i;
  double le = 0, ri = dis[n - 1] - dis[0];
  int ans[3];
  while (le + eps < ri) {
    double mid = (le + ri) * 0.5;
    for (i = 0; i < n; i++)
      if (dis[i] - dis[0] > mid * 2.0) {
        ans[0] = i - 1;
        break;
      }
    if (i == n) {
      ri = mid;
      continue;
    }
    int j = i;
    for (i = i + 1; i < n; i++)
      if (dis[i] - dis[j] > mid * 2.0) {
        ans[1] = i - 1;
        break;
      }
    if (i == n) {
      ri = mid;
      continue;
    }
    if (dis[n - 1] - dis[i] > mid * 2.0)
      le = mid;
    else
      ri = mid;
  }
  le = ri;
  for (i = 0; i < n; i++)
    if (dis[i] - dis[0] > le * 2.0) {
      break;
    }
  double tmp;
  if (i == n) i--;
  tmp = dis[i] + le;
  if (tmp > dis[n - 1] - le) tmp = dis[n - 1] - le;
  printf("%lf\n", le);
  printf("%lf %lf %lf\n", dis[0] + le, tmp, dis[n - 1] - le);
}
void init() {
  for (int i = 0; i < n; i++) scanf("%d", &dis[i]);
}
int main() {
  scanf("%d", &n);
  init();
  work();
  return 0;
}