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 maxn = 210000;
const int inf = 1000000100;
int point[maxn];
int maxl;
int n;
int minl;
int find(int pos) {
  int l = 0;
  int r = n - 1;
  int mid;
  int ok = -1;
  while (l <= r) {
    mid = (l + r) / 2;
    if (point[mid] <= pos) {
      ok = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return ok;
}
int q2, q3;
bool check(int mid) {
  q2 = 0;
  q3 = 0;
  int z = minl + mid;
  int k = find(z);
  if (k == n - 1) return 1;
  q2 = k + 1;
  z = point[k + 1] + mid;
  k = find(z);
  if (k == n - 1) return 1;
  q3 = k + 1;
  z = point[k + 1] + mid;
  if (z >= maxl) return 1;
  return 0;
}
int solve() {
  int l = 0, r = maxl;
  int mid;
  int ok;
  while (l <= r) {
    mid = (l + r) / 2;
    if (check(mid)) {
      ok = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  return ok;
}
int main(int argc, char** argv) {
  scanf("%d", &n);
  maxl = 0;
  minl = inf;
  for (int i = 0; i < n; i++) {
    scanf("%d", &point[i]);
    if (point[i] > maxl) maxl = point[i];
    if (point[i] < minl) minl = point[i];
  }
  sort(point, point + n);
  int d = solve();
  check(d);
  double r = d / 2.0;
  printf("%6lf\n", r);
  printf("%.6lf %.6lf %.6lf\n", point[0] + r, point[q2] + r, point[q3] + r);
  return 0;
}