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;
namespace IO {
const int MAXIOSIZE = 1 << 24 | 1;
unsigned char buf[MAXIOSIZE], *p1, *p2;
template <typename T>
void read(T& x) {
  x = 0;
  char ch =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 24, stdin), p1 == p2)
           ? EOF
           : *p1++);
  bool flg = false;
  for (; ch < '0' || '9' < ch;
       ch = (p1 == p2 &&
                     (p2 = (p1 = buf) + fread(buf, 1, 1 << 24, stdin), p1 == p2)
                 ? EOF
                 : *p1++))
    if (ch == '-') flg |= true;
  for (; '0' <= ch && ch <= '9';
       ch = (p1 == p2 &&
                     (p2 = (p1 = buf) + fread(buf, 1, 1 << 24, stdin), p1 == p2)
                 ? EOF
                 : *p1++))
    x = x * 10 + ch - '0';
  flg ? x = -x : 0;
}
template <typename T>
void out(const T& x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
inline void write(const T& x, const char& ed = ' ') {
  if (x < 0)
    putchar('-'), out(-x);
  else
    out(x);
  putchar(ed);
}
}  // namespace IO
const int MAXN = 2e5 + 10;
const long double eps = 1e-8;
int n;
long double arr[MAXN];
bool chk(long double v, bool out) {
  long double l = arr[n] - 2 * v, r = arr[1] + 2 * v;
  if (l <= r - eps) {
    if (out) {
      printf("%.8Lf %.8Lf 0\n", l, r);
    }
    return true;
  }
  long double l2 = -1, res = -1;
  for (int i = (1), iend = (n); i <= iend; ++i) {
    if (arr[i] >= r + eps) {
      l2 = arr[i] + 2 * v, res = arr[i] + v;
      break;
    }
  }
  if (l2 == -1) {
    printf("%.8Lf %.8Lf 0\n", l, r);
    return true;
  }
  for (int i = (n), iend = (1); i >= iend; --i) {
    if (arr[i] <= l - eps && arr[i] >= l2 + eps) return false;
  }
  if (out) {
    printf("%.8Lf %.8Lf %.8Lf\n", arr[1] + v, arr[n] - v, res);
    return false;
  } else
    return true;
}
int main() {
  IO::read(n);
  if (n <= 3) {
    puts("0");
    for (int i = (1), iend = (n); i <= iend; ++i) {
      int x;
      IO::read(x);
      IO::write(x);
    }
    for (int j = (n + 1), jend = (3); j <= jend; ++j) IO::write(0);
    return 0;
  }
  for (int i = (1), iend = (n); i <= iend; ++i) {
    int x;
    IO::read(x);
    arr[i] = 1. * x;
  }
  sort(arr + 1, arr + n + 1);
  long double l = 0, r = arr[n] - arr[1];
  while (r > l + eps) {
    long double mid = (l + r) / 2;
    if (chk(mid, false))
      r = mid;
    else
      l = mid;
  }
  printf("%.8Lf\n", r);
  chk(r, true);
  return 0;
}