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, point;
set<double> s;
bool check(double x, bool fl) {
  int i = 0, k, j;
  int cnt = 0;
  for (k = 0; k < 3; k++) {
    double base = v[i] + x;
    for (j = i; j < v.size(); j++) {
      if (v[j] <= base) continue;
      if ((v[j] - base - x) <= 0)
        continue;
      else {
        cnt++;
        i = j;
        break;
      }
    }
    if (j < v.size() && k == 2 && !fl) {
      point.clear();
      return false;
    }
    point.push_back(base);
  }
  return true;
}
void make(double r) {
  sort(v.begin(), v.end());
  point.clear();
  int i = 0, j, tmp = i;
  double b1 = v[0] + r;
  point.push_back(v[0] + r);
  i = upper_bound(v.begin(), v.end(), b1 + r) - v.begin();
  if (fabs(v[i] - b1 - r) < 1e-12 && i < v.size() - 1) i += 1;
  if (tmp == i) i += 1;
  b1 = v[i] + r;
  point.push_back(b1);
  tmp = i;
  i = upper_bound(v.begin(), v.end(), b1 + r) - v.begin();
  if (fabs(v[i] - b1 - r) < 1e-12 && i < v.size() - 1) i += 1;
  if (tmp == i) i += 1;
  b1 = v[i] + r;
  point.push_back(b1);
}
double fn() {
  double l = 0, h = 2e9, m;
  for (long long i = 0; i < 300; i++) {
    m = (l + h) / 2.0;
    if (fabs(h - l) < 1e-12) {
      break;
    };
    if (check(m, false)) {
      point.clear();
      h = m;
    } else
      l = m;
  }
  make(m);
  return m;
}
int main() {
  int t, tc = 0;
  long long n;
  while (1 == scanf("%lld", &n)) {
    double x, ans;
    for (long long i = 0; i < n; ++i) {
      scanf("%lf", &x);
      s.insert(x);
    }
    v = vector<double>(s.begin(), s.end());
    if (true)
      ans = fn();
    else {
      puts("0.000000");
      for (auto i : s) printf("%.6f ", (double)i);
      puts("");
      s.clear();
      continue;
    }
    printf("%.6f\n", ans);
    ++tc;
    for (int i = 0; i < point.size(); i++) printf("%.6f ", point[i]);
    if (point.size() < 3)
      for (int i = 0; i < 3 - point.size(); i++) printf("%.6f ", point[0]);
    puts("");
    s.clear();
  }
  return 0;
}