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;
int n;
int pos[210000];
double minim, maxim;
double ansmid;
bool check(double mid) {
  int left = 0, right = 0;
  for (int i = 0; i < n; ++i) {
    if (minim + mid * 2 >= (double)pos[i]) {
      left++;
    } else
      break;
  }
  for (int i = n - 1; i >= 0; --i) {
    if (maxim - mid * 2 <= (double)pos[i]) {
      right++;
    } else
      break;
  }
  double di = (pos[n - right - 1] - pos[left]);
  di /= 2;
  ansmid = pos[left] + di;
  if (left + right >= n || di <= 0.0 || di <= mid) return 1;
  return 0;
}
void solve() {
  minim = 1e+10, maxim = 0;
  ansmid = -1.0;
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &pos[i]);
    if (minim > pos[i]) minim = pos[i];
    if (maxim < pos[i]) maxim = pos[i];
  }
  sort(pos, pos + n);
  if (n <= 3) {
    double dpos[3];
    for (int i = 0; i < (n); ++i) dpos[i] = pos[i];
    if (n == 3) {
      printf("%.f\n", 0.0);
      printf("%.6f %.6f %.6f\n", dpos[0], dpos[1], dpos[2]);
    }
    if (n == 2) {
      printf("%.f\n", 0.0);
      printf("%.6f %.6f %.6f\n", dpos[0], dpos[1], dpos[0] + dpos[1]);
    }
    if (n == 1) {
      printf("%.f\n", 0.0);
      printf("%.6f %.6f %.6f\n", dpos[0], dpos[0] * 2, dpos[0] * 3);
    }
    return;
  }
  double left = 0, right = 1e+10, d = 1e+10;
  double mid, base1, base2, base3;
  for (int i = 0; i < 1000; ++i) {
    mid = (left + right) * 0.5;
    if (check(mid)) {
      if (d > mid) d = mid;
      right = mid;
    } else {
      left = mid;
    }
  }
  printf("%.6f\n", d);
  printf("%.6f %.6f %.6f\n", minim + d, ansmid, maxim - d);
}
int main() {
  while (cin >> n) {
    solve();
  }
}