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-9;
double x[200055];
double ans[3];
int n;
int dmp(double k) {
  if (fabs(k) < EPS) return 0;
  return k < 0 ? -1 : 1;
}
bool check(double r) {
  ans[0] = x[0] + r;
  int idx = 0;
  double ed = x[0] + 2 * r;
  while (idx < n && dmp(x[idx] - ed) <= 0) idx++;
  if (idx == n) {
    ans[2] = ans[1] = ans[0];
    return 1;
  }
  ans[1] = x[idx] + r, ed = x[idx] + 2 * r;
  while (idx < n && dmp(x[idx] - ed) <= 0) idx++;
  if (idx == n) {
    ans[2] = ans[1];
    return 1;
  }
  ans[2] = x[idx] + r, ed = x[idx] + 2 * r;
  while (idx < n && dmp(x[idx] - ed) <= 0) idx++;
  return idx == n;
}
int main() {
  while (scanf("%d", &n) == 1) {
    for (int i = 0; i < n; i++) scanf("%lf", x + i);
    sort(x, x + n);
    double l = 0, r = x[n - 1] - x[0];
    int sum = 0;
    while (l < r && sum < 200) {
      double mid = (l + r) / 2;
      if (check(mid))
        r = mid;
      else
        l = mid + EPS;
      sum++;
    }
    if (fabs(r) > EPS)
      printf("%.6f\n", r);
    else
      printf("%d\n", 0);
    check(r);
    printf("%.6f %.6f %.6f\n", ans[0], ans[1], ans[2]);
  }
  return 0;
}