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 max3(int a, int b, int c) {
  if (a > b) {
    if (c > a)
      return c;
    else
      return a;
  } else {
    if (c > b)
      return c;
    else
      return b;
  }
}
int main() {
  int n, i, b, e, x1, y1, x2, y2, x3, y3, g1, g2, m = 2000000000, j, x;
  vector<int> v;
  scanf("%d", &n);
  v.resize(n);
  for (i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  sort(v.begin(), v.end());
  if (n < 4) {
    printf("0\n");
    while (v.size() < 3) {
      v.push_back(v[v.size() - 1]);
    }
    for (i = 0; i < 3; i++) {
      printf("%d.000000 ", v[i]);
    }
    return 0;
  }
  for (i = 0; i < n - 2; i++) {
    b = i + 1;
    e = n - 2;
    while (e - b > 1) {
      x1 = v[b] - v[i + 1];
      y1 = v[n - 1] - v[b + 1];
      x2 = v[e] - v[i + 1];
      y2 = v[n - 1] - v[e + 1];
      x1 = y1 - x1;
      x2 = y2 - x2;
      if (x1 == 0) {
        e = b;
        break;
      }
      if (x2 == 0) {
        b = e;
        break;
      }
      if (x1 > 0 && x2 > 0) {
        b = (e + b) / 2;
        continue;
      }
      if (x1 < 0 && x2 < 0) {
        e = (e + b) / 2;
        continue;
      }
      x = (e + b) / 2;
      x3 = v[x] - v[i + 1];
      y3 = v[n - 1] - v[x + 1];
      x3 = y3 - x3;
      if (x3 == 0) {
        e = b = x;
        break;
      }
      if (x3 > 0) {
        b = x;
      } else {
        e = x;
      }
    }
    x1 = max(v[b] - v[i + 1], v[n - 1] - v[b + 1]);
    y1 = b;
    for (j = b + 1; j <= e; j++) {
      x2 = max(v[j] - v[i + 1], v[n - 1] - v[j + 1]);
      if (x2 < x1) {
        x1 = x2;
        y1 = j;
      }
    }
    e = y1;
    b = max3(v[i] - v[0], v[e] - v[i + 1], v[n - 1] - v[e + 1]);
    if (b < m) {
      g1 = i;
      g2 = e;
      m = b;
    }
  }
  printf("%d.%d00000\n%d.%d00000 %d.%d00000 %d.%d00000", m / 2, m % 2 * 5,
         (v[g1] + v[0]) / 2, (v[g1] + v[0]) % 2 * 5, (v[g2] + v[g1 + 1]) / 2,
         (v[g2] + v[g1 + 1]) % 2 * 5, (v[n - 1] + v[g2 + 1]) / 2,
         (v[n - 1] + v[g2 + 1]) % 2 * 5);
  return 0;
}