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 loc[200000 + 10];
int n;
int ans[4];
int fnext(int y) {
  int l = 1, r = n + 1, mid = (l + r) / 2;
  while (l < r) {
    if (loc[mid] <= y)
      l = mid + 1;
    else
      r = mid;
    mid = (l + r) / 2;
  }
  return mid;
}
int ok(int x) {
  int i, j;
  int z = 1;
  for (i = 1; i <= 3; i++) {
    z = fnext(loc[z] + x);
    ans[i] = z;
    if (z >= n + 1) return 1;
  }
  return 0;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    int i, j, k;
    for (i = 1; i <= n; i++) {
      scanf("%d", &loc[i]);
    }
    sort(loc + 1, loc + n + 1);
    int r = (loc[n] - loc[1]);
    int l = 0, mid;
    while (l < r) {
      mid = (r + l) / 2;
      if (ok(mid))
        r = mid;
      else
        l = mid + 1;
    }
    ok(l);
    double radius = l * 1.0 / 2.0;
    double radar1 = (loc[ans[1] - 1] + loc[1]) * 1.0 / 2.0;
    double radar2 = (loc[ans[2] - 1] + loc[ans[1]]) / 2.0;
    double radar3 = (loc[ans[3] - 1] + loc[ans[2]]) / 2.0;
    printf("%.6f\n", radius);
    printf("%.6f ", radar1);
    printf("%.6f ", radar2);
    printf("%.6f\n", radar3);
  }
  return 0;
}