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;
double const pi = 3.1415926535897932384626433832795;
long long const inf64 = (long long)2e18;
int n;
long long x[200001], p[4];
bool may(long long r) {
  long long last = (long long)-1e15;
  int cnt = 0;
  for (int i = 0; i < (int)n; i++)
    if (last + r < x[i]) last = x[i] + r, p[cnt++] = x[i] + r;
  while (cnt < 3) {
    p[cnt + 1] = p[cnt];
    cnt++;
  }
  if (cnt <= 3)
    return true;
  else
    return false;
}
int main() {
  cin >> n;
  for (int i = 0; i < (int)n; i++) {
    int tmp;
    scanf("%d", &tmp);
    x[i] = (long long)tmp * 2;
  }
  sort(x, x + n);
  long long l = 0, r = (long long)1e10;
  while (r - l > 1) {
    long long m = (l + r) / 2;
    if (may(m))
      r = m;
    else
      l = m;
  }
  long long ans;
  if (may(l))
    ans = l;
  else {
    may(r);
    ans = r;
  }
  printf("%.6lf\n", (double)ans / 2);
  for (int i = 0; i < (int)3; i++) printf("%.6lf ", (double)p[i] / 2);
  return 0;
}