#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f;
const double inf = 1e18;
const double eps = 1e-9;
const long long mod = 998244353;
const unsigned long long mx = 133333331;
inline void RI(int &x) {
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0';
}
int a[300010];
double p[3];
int n;
bool judge(double d) {
  double cnt = -1;
  int nt = 0;
  double k[3];
  for (int i = 1; i <= n; i++) {
    if (a[i] > cnt) {
      if (nt == 3) {
        return false;
      }
      k[nt++] = a[i] + d;
      cnt = a[i] + 2 * d;
    }
  }
  for (int i = nt; i < 3; i++) k[i] = 0;
  for (int i = 0; i < 3; i++) p[i] = k[i];
  return true;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  sort(a + 1, a + n + 1);
  double l = 0, r = 2e9 + 1;
  double ans = 0;
  int num = 200;
  while (r - l > 1e-7 && num != 0) {
    double m = (l + r) / 2.0;
    int res = judge(m);
    if (res == 1) {
      ans = m;
      r = m;
    } else
      l = m;
    num--;
  }
  printf("%.6f\n", ans);
  for (int i = 0; i < 3; i++) {
    printf("%.6f", p[i]);
    if (i == 2)
      printf("\n");
    else
      printf(" ");
  }
}

### Prompt

Please formulate 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;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f;
const double inf = 1e18;
const double eps = 1e-9;
const long long mod = 998244353;
const unsigned long long mx = 133333331;
inline void RI(int &x) {
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0';
}
int a[300010];
double p[3];
int n;
bool judge(double d) {
  double cnt = -1;
  int nt = 0;
  double k[3];
  for (int i = 1; i <= n; i++) {
    if (a[i] > cnt) {
      if (nt == 3) {
        return false;
      }
      k[nt++] = a[i] + d;
      cnt = a[i] + 2 * d;
    }
  }
  for (int i = nt; i < 3; i++) k[i] = 0;
  for (int i = 0; i < 3; i++) p[i] = k[i];
  return true;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  sort(a + 1, a + n + 1);
  double l = 0, r = 2e9 + 1;
  double ans = 0;
  int num = 200;
  while (r - l > 1e-7 && num != 0) {
    double m = (l + r) / 2.0;
    int res = judge(m);
    if (res == 1) {
      ans = m;
      r = m;
    } else
      l = m;
    num--;
  }
  printf("%.6f\n", ans);
  for (int i = 0; i < 3; i++) {
    printf("%.6f", p[i]);
    if (i == 2)
      printf("\n");
    else
      printf(" ");
  }
}

```