#include <bits/stdc++.h>
using namespace std;
double mm[200005];
int n;
double ans[6];
bool work(double aa) {
  int sp, i;
  ans[0] = mm[0] + aa;
  sp = 1;
  for (i = 1; i < n; i++) {
    if (fabs(mm[i] - ans[sp - 1]) - aa <= 0.0000001) continue;
    ans[sp++] = mm[i] + aa;
    if (sp == 4) break;
  }
  if (sp == 4) return 0;
  return 1;
}
int main() {
  int i;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> mm[i];
  }
  sort(mm, mm + n);
  if (n <= 3) {
    printf("0.000000\n");
    if (n == 3) {
      printf("%lf %lf %lf\n", mm[0], mm[1], mm[2]);
    }
    if (n == 2) {
      printf("%lf %lf %lf\n", mm[0], mm[1], mm[1]);
    }
    if (n == 1) {
      printf("%lf %lf %lf\n", mm[0], mm[0], mm[0]);
    }
    return 0;
  }
  double mmax, mmin, mmid;
  mmin = 0;
  mmax = (mm[n - 1] - mm[0]) / 3;
  mmid = (mmax + mmin) / 2;
  while (mmax - mmin > 0.0000001) {
    mmid = (mmax + mmin) / 2;
    if (work(mmid)) {
      mmax = mmid;
    } else
      mmin = mmid;
  }
  if (work(mmid)) {
    printf("%lf\n", mmid);
    printf("%lf %lf %lf\n", ans[0], ans[1], ans[2]);
  } else {
    work(mmax);
    printf("%lf\n", mmax);
    printf("%lf %lf %lf\n", ans[0], ans[1], ans[2]);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
using namespace std;
double mm[200005];
int n;
double ans[6];
bool work(double aa) {
  int sp, i;
  ans[0] = mm[0] + aa;
  sp = 1;
  for (i = 1; i < n; i++) {
    if (fabs(mm[i] - ans[sp - 1]) - aa <= 0.0000001) continue;
    ans[sp++] = mm[i] + aa;
    if (sp == 4) break;
  }
  if (sp == 4) return 0;
  return 1;
}
int main() {
  int i;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> mm[i];
  }
  sort(mm, mm + n);
  if (n <= 3) {
    printf("0.000000\n");
    if (n == 3) {
      printf("%lf %lf %lf\n", mm[0], mm[1], mm[2]);
    }
    if (n == 2) {
      printf("%lf %lf %lf\n", mm[0], mm[1], mm[1]);
    }
    if (n == 1) {
      printf("%lf %lf %lf\n", mm[0], mm[0], mm[0]);
    }
    return 0;
  }
  double mmax, mmin, mmid;
  mmin = 0;
  mmax = (mm[n - 1] - mm[0]) / 3;
  mmid = (mmax + mmin) / 2;
  while (mmax - mmin > 0.0000001) {
    mmid = (mmax + mmin) / 2;
    if (work(mmid)) {
      mmax = mmid;
    } else
      mmin = mmid;
  }
  if (work(mmid)) {
    printf("%lf\n", mmid);
    printf("%lf %lf %lf\n", ans[0], ans[1], ans[2]);
  } else {
    work(mmax);
    printf("%lf\n", mmax);
    printf("%lf %lf %lf\n", ans[0], ans[1], ans[2]);
  }
  return 0;
}

```