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 X = 200000 + 5;
double v[X];
int n;
int nxt(int st, double m) {
  for (int i = st; i < n; i++) {
    if (v[i] <= v[st] + 2.0 * m + 1e-9) {
    } else {
      return i;
    }
  }
  return n;
}
bool can(double m) {
  int p1 = nxt(0, m);
  if (p1 == n) return 1;
  p1 = nxt(p1, m);
  if (p1 == n) return 1;
  p1 = nxt(p1, m);
  if (p1 == n) return 1;
  return 0;
}
void F(double m) {
  printf("%.6f ", v[0] + m);
  int p1 = nxt(0, m);
  if (p1 == n) p1 = 0;
  printf("%.6f ", v[p1] + m);
  p1 = nxt(p1, m);
  if (p1 == n) p1 = 0;
  printf("%.6f ", v[p1] + m);
}
void MAIN() {
  long long pos = 0, m;
  cin >> n;
  m = n;
  while (n--) {
    double x;
    cin >> x;
    v[pos] = x;
    pos++;
  }
  n = m;
  sort(v, v + m);
  double lo = 0, hi = 1e10;
  for (int i = 0; i < 200; i++) {
    double m = (lo + hi) / 2.0;
    if (can(m))
      hi = m;
    else
      lo = m;
  }
  printf("%.6f\n", hi);
  F(hi);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << setprecision(12);
  MAIN();
  return 0;
}