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;
struct Point {
  int x;
  int y;
  Point(int x_, int y_) : x(x_), y(y_) {}
};
template <class T>
T sqr(const T& t) {
  return t * t;
}
template <class T>
T abs(const T& t) {
  return ((t > 0) ? (t) : (-t));
}
void initialize() {
  freopen("_.in", "r", stdin);
  freopen("_.out", "w", stdout);
}
const int MAX = (int)2e5 + 10;
long long a[MAX];
long long res[3];
int n;
bool can(long long mid) {
  long long cur = a[0];
  int st = 0;
  int pos = 0;
  while (st < 3) {
    pos = upper_bound(a, a + n, cur + 2 * mid) - a;
    res[st] = cur + mid;
    cur = a[pos];
    ++st;
    if (pos == n) break;
  }
  while (st < 3) {
    res[st] = res[st - 1];
    st += 1;
  }
  if (pos == n) return true;
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
    a[i] *= 2;
  }
  sort(a, a + n);
  long long down = 0, up = (long long)2e9;
  while (up - down > 1) {
    long long mid = (down + up) / 2.0;
    if (can(mid)) {
      up = mid;
    } else {
      down = mid;
    }
  }
  long long r;
  if (can(down)) {
    r = down;
  } else {
    can(up);
    r = up;
  }
  printf("%.6lf\n", (double)r / 2.0);
  for (int i = 0; i < 3; ++i) {
    printf("%.6lf ", (double)res[i] / 2.0);
  }
  return 0;
}