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>
int N;
double X[202020], a, b, c, p, A, B, C;
bool CHK(double d) {
  int i, x = 0;
  a = X[x] + d;
  for (i = x + 1; i < N; i++) {
    if (X[i] > a + d) break;
  }
  x = i;
  if (x == N) {
    b = a + d * 2 + 1;
    c = b + d * 2 + 1;
    return true;
  }
  b = X[x] + d;
  for (i = x + 1; i < N; i++) {
    if (X[i] > b + d) break;
  }
  x = i;
  if (x == N) {
    c = b + d * 2 + 1;
    return true;
  }
  c = X[x] + d;
  if (X[N - 1] > c + d) return false;
  return true;
}
int main() {
  int i;
  scanf("%d", &N);
  for (i = 0; i < N; i++) scanf("%lf", &X[i]);
  std::sort(X, X + N);
  double l = 0, r = 3e9;
  double m;
  p = 3e9;
  for (i = 0; i < 150; i++) {
    m = double(l + r) / 2;
    if (CHK(m)) {
      r = m;
      if (p > m) p = m, A = a, B = b, C = c;
    } else
      l = m;
  }
  printf("%.6lf\n%.6lf %.6lf %.6lf", p, A, B, C);
  return 0;
}