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;
template <typename T>
int size(const T& c) {
  return int(c.size());
}
template <typename T>
T abs(T x) {
  return x < 0 ? -x : x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
bool remin(T& x, T y) {
  if (x <= y) return false;
  x = y;
  return true;
}
template <typename T>
bool remax(T& x, T y) {
  if (x >= y) return false;
  x = y;
  return true;
}
int n;
vector<long long> x;
bool tryPut(long long d, long long& x1, long long& x2, long long& x3) {
  int i = 0, c = 0;
  while (i < n && c < 3) {
    ++c;
    if (c == 1)
      x1 = x[i];
    else if (c == 2)
      x2 = x[i];
    else
      x3 = x[i];
    long long from = x[i];
    while (i < n && x[i] <= from + d) ++i;
  }
  if (i < n) return false;
  if (c == 1) {
    x2 = x3 = x1;
  } else if (c == 2) {
    x3 = x2;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  x.resize(n);
  for (int i(0), _n(n); i < _n; ++i) scanf("%lld", &x[i]);
  sort(x.begin(), x.end());
  long long ans, x1, x2, x3;
  if (tryPut(0, x1, x2, x3))
    ans = 0;
  else {
    int L = 0, R = 1;
    while (!tryPut(R, x1, x2, x3)) {
      L = R;
      R *= 2;
    }
    while (R - L > 1) {
      int M = (L + R) / 2;
      if (tryPut(M, x1, x2, x3))
        R = M;
      else
        L = M;
    }
    tryPut(R, x1, x2, x3);
    ans = R;
  }
  printf("%.6lf\n%.6lf %.6lf %.6lf\n", 0.5 * ans, x1 + 0.5 * ans,
         x2 + 0.5 * ans, x3 + 0.5 * ans);
}