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>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
inline int read() {
  int x = 0;
  char c = getchar();
  bool k = false;
  for (; !isdigit(c); c = getchar()) k |= c == '-';
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  if (k) x = -x;
  return x;
}
void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 2e5 + 100;
int a[maxn], n;
bool check(double d) {
  double t1 = a[1] + 2 * d;
  double t2 = a[n] - 2 * d;
  int n1 = *upper_bound(a + 1, a + 1 + n, t1);
  int n2 = *(lower_bound(a + 1, a + 1 + n, t2) - 1);
  if (n1 + 2 * d >= n2)
    return true;
  else
    return false;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + 1 + n);
  n = unique(a + 1, a + 1 + n) - a - 1;
  double l = 0, r = (a[n] - a[1]) * 0.5;
  double ans = 0;
  while (l + 1e-7 < r) {
    double mid = (l + r) / 2;
    if (check(mid)) {
      ans = mid;
      r = mid;
    } else {
      l = mid;
    }
  }
  double d = ans;
  printf("%.6lf\n", d);
  double t1 = a[1] + 2 * d;
  double t2 = a[n] - 2 * d;
  int n1 = *upper_bound(a + 1, a + 1 + n, t1);
  printf("%.6lf %.6lf %.6lf", a[1] + d, n1 + d, a[n] - d);
}