#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int f_min(int x, int y) { return x < y ? x : y; }
int f_max(int x, int y) { return x > y ? x : y; }
void f_swap(int &x, int &y) {
  int t;
  t = x, x = y, y = t;
}
int f_abs(int x) { return x > 0 ? x : -x; }
int a[2001001];
int n;
double x[10];
int f(int val) {
  int l = 0, r = n - 1, mid;
  mid = 0;
  if (l == r) return r + 1;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (a[mid] > val)
      r = mid;
    else
      l = mid;
  }
  if (a[r] <= val) return r + 1;
  if (a[l] > val) return l;
  if (a[mid] > val) return mid;
  return r;
}
bool run(int val) {
  int i;
  int t = 0;
  x[0] = 0, x[1] = 0, x[2] = 0;
  for (i = 0; i < 3; i++) {
    x[i] = a[t] + val * 1.0 / 2;
    t = f(a[t] + val);
    if (t >= n) return true;
  }
  return false;
}
int find() {
  int l = 0, r = 2000000000, mid;
  while (r - l > 1) {
    mid = (r + l) / 2;
    if (run(mid))
      r = mid;
    else
      l = mid;
  }
  if (run(l)) return l;
  if (run(mid)) return mid;
  return r;
}
int main() {
  int i;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) scanf("%d", &a[i]);
    sort(a, a + n);
    int ans = find();
    run(ans);
    printf("%.6lf\n%.6lf %.6lf %.6lf\n", ans * 1.0 / 2, x[0], x[1], x[2]);
  }
  return 0;
}

### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int f_min(int x, int y) { return x < y ? x : y; }
int f_max(int x, int y) { return x > y ? x : y; }
void f_swap(int &x, int &y) {
  int t;
  t = x, x = y, y = t;
}
int f_abs(int x) { return x > 0 ? x : -x; }
int a[2001001];
int n;
double x[10];
int f(int val) {
  int l = 0, r = n - 1, mid;
  mid = 0;
  if (l == r) return r + 1;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (a[mid] > val)
      r = mid;
    else
      l = mid;
  }
  if (a[r] <= val) return r + 1;
  if (a[l] > val) return l;
  if (a[mid] > val) return mid;
  return r;
}
bool run(int val) {
  int i;
  int t = 0;
  x[0] = 0, x[1] = 0, x[2] = 0;
  for (i = 0; i < 3; i++) {
    x[i] = a[t] + val * 1.0 / 2;
    t = f(a[t] + val);
    if (t >= n) return true;
  }
  return false;
}
int find() {
  int l = 0, r = 2000000000, mid;
  while (r - l > 1) {
    mid = (r + l) / 2;
    if (run(mid))
      r = mid;
    else
      l = mid;
  }
  if (run(l)) return l;
  if (run(mid)) return mid;
  return r;
}
int main() {
  int i;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) scanf("%d", &a[i]);
    sort(a, a + n);
    int ans = find();
    run(ans);
    printf("%.6lf\n%.6lf %.6lf %.6lf\n", ans * 1.0 / 2, x[0], x[1], x[2]);
  }
  return 0;
}

```