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;
int n;
int a[200009];
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    sort(a, a + n);
    int l = 0, r = a[n - 1] - a[0];
    int p;
    while (l < r) {
      int mid = (r + l) / 2;
      p = 0;
      for (int i = 0; i < 3; i++) p = upper_bound(a + p, a + n, a[p] + mid) - a;
      if (p < n)
        l = mid + 1;
      else
        r = mid;
    }
    printf("%d.%c00000\n", r / 2, r & 1 ? '5' : '0');
    p = 0;
    for (int i = 0; i < 3; i++) {
      printf("%d.%c00000 ", (p < n ? a[p] : 0) + r / 2, r & 1 ? '5' : '0');
      p = upper_bound(a + p, a + n, a[p] + r) - a;
    }
  }
  return 0;
}