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 MAXN = 200000 + 100;
int x[MAXN];
int n;
int getnext(double dis) {
  if (dis >= x[n - 1]) return n;
  int down = 0, up = n - 1;
  while (down + 1 < up) {
    int mid = (down + up) >> 1;
    if (x[mid] > dis) {
      up = mid;
    } else {
      down = mid;
    }
  }
  return up;
}
int getlast(double dis) {
  if (dis <= x[0]) return -1;
  int down = 0, up = n - 1;
  while (down + 1 < up) {
    int mid = (down + up) >> 1;
    if (x[mid] < dis) {
      down = mid;
    } else {
      up = mid;
    }
  }
  return down;
}
bool isok(double r) {
  double xx = x[0] + r + r;
  int ind = getnext(xx);
  if (ind == n) return true;
  xx = x[n - 1] - r - r;
  int ind2 = getlast(xx);
  if (ind2 == -1) return true;
  if (ind2 < ind) return true;
  return x[ind2] - x[ind] < r + r;
}
void gao(double r) {
  printf("%.6lf\n", r);
  int ind = getnext(x[0] + r + r);
  int ind2 = getlast(x[n - 1] - r - r);
  double mid;
  if (ind2 < ind) {
    mid = x[ind];
  } else {
    mid = (x[ind2] + x[ind]) / 2.0;
  }
  printf("%.6lf %.6lf %.6lf\n", x[0] + r, mid, x[n - 1] - r);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", x + i);
  }
  sort(x, x + n);
  double down = 0, up = (x[n - 1] - x[0]) / 5.9;
  for (int i = 0; i < 100; ++i) {
    double mid = (down + up) / 2;
    if (isok(mid)) {
      up = mid;
    } else {
      down = mid;
    }
  }
  gao(up);
  return 0;
}