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;
vector<double> v;
bool ok(double mid, bool f) {
  double pos = mid + v[0];
  int indx = upper_bound(v.begin(), v.end(), pos + mid) - v.begin();
  if (f) {
    if (pos > v[v.size() - 1])
      printf("%.6lf ", v[v.size() - 1]);
    else
      printf("%.6lf ", pos);
  }
  if (indx < v.size()) {
    pos = v[indx] + mid;
    indx = upper_bound(v.begin(), v.end(), pos + mid) - v.begin();
  }
  if (f) {
    if (pos > v[v.size() - 1])
      printf("%.6lf", v[v.size() - 1]);
    else
      printf("%.6lf ", pos);
  }
  if (indx < v.size()) {
    pos = v[indx] + mid;
    indx = upper_bound(v.begin(), v.end(), pos + mid) - v.begin();
  }
  if (f) {
    if (pos > v[v.size() - 1])
      printf("%.6lf", v[v.size() - 1]);
    else
      printf("%.6lf ", pos);
  }
  if (pos + mid >= v[v.size() - 1]) {
    return true;
  } else
    return false;
}
double bs() {
  double lo = 0, hi = 1000000000, mid;
  for (int i = 0; i < 100; i++) {
    mid = (lo + hi) / 2;
    if (ok(mid, false)) {
      hi = mid;
    } else
      lo = mid;
  }
  return hi;
}
int main() {
  int n;
  double num, mid;
  scanf("%d", &n);
  while (n--) {
    scanf("%lf", &num);
    v.push_back(num);
  }
  sort(v.begin(), v.end());
  mid = bs();
  printf("%.6lf\n", mid);
  ok(mid, true);
  return 0;
}