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>
int f_min(int x, int y) {
  if (x > y)
    return y;
  else
    return x;
}
int f_max(int x, int y) {
  if (x > y)
    return x;
  else
    return y;
}
int f_abs(int x) { return (x) > (0) ? (x) : (-x); }
void swap(int& x, int& y) {
  int t;
  t = x;
  x = y;
  y = t;
}
int g = 1;
void printcase() { printf("PUZZLE #%d\n", g++); }
using namespace std;
const int MM = 511111;
const double esp = 1e-8;
const double lep = 0.0000001;
int N, M, mm;
int x[MM];
double xx[MM];
void get_data() {
  int i, j, k;
  for (i = 0; i < N; i++) {
    scanf("%d", &x[i]);
  }
  sort(x, x + N);
  mm = 1;
  for (i = 1; i < N; i++)
    if (x[i] != x[i - 1]) x[mm++] = x[i];
  N = mm;
  for (i = 0; i < N; i++) xx[i] = (double)(1.0 * x[i]);
}
int find(double val) {
  int l = 0, r = N - 1, mid;
  while (l <= r) {
    mid = (l + r) >> 1;
    if ((xx[mid] - val) > esp)
      r = mid - 1;
    else
      l = mid + 1;
  }
  return l;
}
void solve() {
  int i, j, k, id, tt;
  bool ff = false;
  double ans, tmp, sum, d, a0 = 0.00, a1 = 0.00, a2 = 0.00;
  double l = 0, r = xx[N - 1], mid;
  while ((r - l) > esp) {
    d = l + r;
    mid = d / 2.0;
    tmp = xx[0] + d;
    a0 = xx[0] + mid;
    if (tmp + esp >= xx[N - 1]) {
      r = mid - esp;
      continue;
    }
    id = find(tmp);
    tmp = xx[id] + d;
    a1 = xx[id] + mid;
    if (tmp + esp >= xx[N - 1]) {
      r = mid - esp;
      continue;
    }
    id = find(tmp);
    if (xx[id] + d + esp >= xx[N - 1]) {
      a2 = xx[id] + mid;
      r = mid - esp;
    } else
      l = mid + esp;
  }
  printf("%lf\n", l);
  a0 = xx[0] + l;
  tmp = xx[0] + (double)(l * 2.0);
  id = find(tmp);
  if (tmp + lep >= xx[id] && id + 1 < N) id++;
  a1 = xx[id] + l;
  tmp = xx[id] + (double)(l * 2.0);
  tt = find(tmp);
  if (tmp + lep >= xx[tt] && tt + 1 < N) tt++;
  a2 = xx[tt] + l;
  printf("%lf %lf %lf\n", a0, a1, a2);
}
int main() {
  while (scanf("%d", &N) != EOF) get_data(), solve();
  return 0;
}