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 mxn = 2e5 + 5;
int n;
long long x[mxn];
long long a, b, c;
bool chk(long long d) {
  int i = 0, j = n - 1;
  while (i < n && x[i] - x[0] <= d * 2) ++i;
  a = (x[i - 1] + x[0]) / 2;
  while (j >= 0 && x[n - 1] - x[j] <= d * 2) --j;
  c = (x[n - 1] + x[j + 1]) / 2;
  if (i <= j && x[j] - x[i] > d * 2) return 0;
  b = (x[j] + x[i]) / 2;
  return 1;
}
inline void P2(long long x) {
  printf("%lld", x >> 1);
  if (x & 1)
    printf(".500000");
  else
    printf(".000000");
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); ++i) scanf("%lld", &x[i]);
  sort(x, x + n);
  for (int i = 0; i < (int)(n); ++i) x[i] <<= 1;
  long long lb = 0, rb = 2e9;
  while (lb < rb) {
    long long md = (lb + rb) >> 1;
    if (chk(md))
      rb = md;
    else
      lb = md + 1;
  }
  chk(lb);
  P2(lb);
  puts("");
  P2(a);
  printf(" ");
  P2(b);
  printf(" ");
  P2(c);
  puts("");
  return 0;
}