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>
using namespace std;
char BUF[1 << 21], *P1 = BUF, *P2 = BUF;
inline int read() {
  int res = 0, f = 1;
  char ch =
      (P1 == P2 && (P2 = (P1 = BUF) + fread(BUF, 1, 1 << 21, stdin), P1 == P2)
           ? EOF
           : *P1++);
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch =
        (P1 == P2 && (P2 = (P1 = BUF) + fread(BUF, 1, 1 << 21, stdin), P1 == P2)
             ? EOF
             : *P1++);
  }
  while (ch >= '0' && ch <= '9')
    res = (res << 3) + (res << 1) + ch - 48,
    ch =
        (P1 == P2 && (P2 = (P1 = BUF) + fread(BUF, 1, 1 << 21, stdin), P1 == P2)
             ? EOF
             : *P1++);
  return res * f;
}
const int N = 2e5 + 100;
const double eps = 1e-7;
int n, a[N];
double l, r = 2e9, d, ans;
bool pd(double mid) {
  double la = a[1];
  int tot = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] - la > 2 * mid) {
      tot++;
      la = a[i];
    }
  }
  if (tot <= 3) return true;
  return false;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  while (l <= r) {
    double mid = (l + r) / 2;
    if (pd(mid))
      r = mid - eps, ans = mid;
    else
      l = mid + eps;
  }
  printf("%.6lf\n", ans);
  double la = a[1];
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] - la > 2 * ans) {
      tot++;
      printf("%.6lf ", (a[i - 1] - la) / 2 + la);
      la = a[i];
    }
  }
  if (tot < 3) printf("%.6lf ", (a[n] - la) / 2 + la), tot++;
  while (tot < 3) printf("%.6lf ", a[n]), tot++;
  return 0;
}