#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
using namespace std;
const double PI = acos(-1.0);
const int MAXN = 5e5 + 55;
const int INF = 0x3f3f3f3f;
const int NINF = 0x80808080;
const long long NLINF = 0x8080808080808080;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
int n;
double a[MAXN];
vector<double> ans;
int check(double d) {
  ans.clear();
  ans.push_back(a[1] + d);
  double now = a[1] + 2 * d;
  for (int i = 1; i <= n; i++) {
    if (a[i] > now) {
      ans.push_back(a[i] + d);
      now = a[i] + 2 * d;
      if (ans.size() > 3) return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
  sort(a + 1, a + 1 + n);
  if (n <= 3) {
    printf("0.000000\n");
    for (int i = 1; i <= n; i++) printf("%.6f ", a[i]);
    int dis = 3 - n;
    for (int i = 1; i <= dis; i++) printf("%.6f ", a[n]);
  } else {
    double l = 0, r = 1e9 + 7;
    while (l < r - eps) {
      double mid = (l + r) / 2.0;
      if (check(mid) == 0)
        l = mid + eps;
      else
        r = mid;
    }
    int ff = check(l);
    while (check(l) == 0) {
      l += eps;
    }
    check(l);
    printf("%.6f\n", l);
    for (auto x : ans) printf("%.6f ", x);
    int dis = 3 - ans.size();
    for (int i = 1; i <= dis; i++) printf("%.6f ", a[n]);
  }
}

### Prompt

In CPP, your task is to solve 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 GCC optimize("O3")
#pragma G++ optimize("O3")
using namespace std;
const double PI = acos(-1.0);
const int MAXN = 5e5 + 55;
const int INF = 0x3f3f3f3f;
const int NINF = 0x80808080;
const long long NLINF = 0x8080808080808080;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
int n;
double a[MAXN];
vector<double> ans;
int check(double d) {
  ans.clear();
  ans.push_back(a[1] + d);
  double now = a[1] + 2 * d;
  for (int i = 1; i <= n; i++) {
    if (a[i] > now) {
      ans.push_back(a[i] + d);
      now = a[i] + 2 * d;
      if (ans.size() > 3) return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
  sort(a + 1, a + 1 + n);
  if (n <= 3) {
    printf("0.000000\n");
    for (int i = 1; i <= n; i++) printf("%.6f ", a[i]);
    int dis = 3 - n;
    for (int i = 1; i <= dis; i++) printf("%.6f ", a[n]);
  } else {
    double l = 0, r = 1e9 + 7;
    while (l < r - eps) {
      double mid = (l + r) / 2.0;
      if (check(mid) == 0)
        l = mid + eps;
      else
        r = mid;
    }
    int ff = check(l);
    while (check(l) == 0) {
      l += eps;
    }
    check(l);
    printf("%.6f\n", l);
    for (auto x : ans) printf("%.6f ", x);
    int dis = 3 - ans.size();
    for (int i = 1; i <= dis; i++) printf("%.6f ", a[n]);
  }
}

```