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;
const int N = 2 * 10e5 + 5;
int a[N], n, ans[4];
bool judge(int m) {
  ans[0] = a[0] + m;
  ans[1] = ans[2] = 2000000000;
  int i, t = 0;
  for (i = 0; i < n; i++) {
    if (t >= 3)
      return false;
    else if (ans[t] < a[i])
      ans[++t] = a[i] + m;
  }
  return t < 3;
}
int main() {
  int i;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) scanf("%d", a + i);
    sort(a, a + n);
    int l = -1, r = a[n - 1] + 1;
    while (l + 1 < r) {
      int m = (l + r) >> 1;
      if (judge(m))
        r = m;
      else
        l = m;
    }
    judge(r);
    printf("%.7lf\n%.7lf %.7lf %.7lf\n", r * 0.5, ans[0] - r * 0.5,
           ans[1] - r * 0.5, ans[2] - r * 0.5);
  }
  return 0;
}