#include <bits/stdc++.h>
using namespace std;
int n;
long double r, rsx[1010], rsy[1010], xx[1010], st, dr, mid, cnt, rr;
long double getdist(long double x, long double y, long double xx,
                    long double yy) {
  return (xx - x) * (xx - x) + (yy - y) * (yy - y);
}
bool check(long double val, int p) {
  for (int i = 1; i <= cnt; i++)
    if (getdist(rsx[i], rsy[i], xx[p], val) - rr <= 0.000000000000001) return 0;
  return 1;
}
int main() {
  cin >> n >> r;
  for (int i = 1; i <= n; i++) cin >> xx[i];
  rsx[1] = xx[1];
  rsy[1] = r;
  rr = 4 * r * r;
  cnt = 1;
  for (int i = 2; i <= n; i++) {
    long double bst = r;
    for (int j = 1; j <= cnt; j++)
      if (abs(xx[i] - rsx[j]) <= 2 * r)
        if (rsy[j] > bst) bst = rsy[j];
    st = bst;
    dr = 4e6;
    for (int j = 1; j <= 80; j++) {
      mid = (st + dr) * 0.5;
      if (check(mid, i))
        dr = mid;
      else
        st = mid;
    }
    rsx[i] = xx[i];
    rsy[i] = (st + dr) * 0.5;
    cnt++;
  }
  for (int i = 1; i <= n; i++)
    cout << fixed << setprecision(10) << rsy[i] << ' ';
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

Carol is currently curling.

She has n disks each with radius r on the 2D plane. 

Initially she has all these disks above the line y = 10100.

She then will slide the disks towards the line y = 0 one by one in order from 1 to n. 

When she slides the i-th disk, she will place its center at the point (xi, 10100). She will then push it so the disk’s y coordinate continuously decreases, and x coordinate stays constant. The disk stops once it touches the line y = 0 or it touches any previous disk. Note that once a disk stops moving, it will not move again, even if hit by another disk. 

Compute the y-coordinates of centers of all the disks after all disks have been pushed.

Input

The first line will contain two integers n and r (1 ≤ n, r ≤ 1 000), the number of disks, and the radius of the disks, respectively.

The next line will contain n integers x1, x2, ..., xn (1 ≤ xi ≤ 1 000) — the x-coordinates of the disks.

Output

Print a single line with n numbers. The i-th number denotes the y-coordinate of the center of the i-th disk. The output will be accepted if it has absolute or relative error at most 10 - 6.

Namely, let's assume that your answer for a particular value of a coordinate is a and the answer of the jury is b. The checker program will consider your answer correct if <image> for all coordinates.

Example

Input

6 2
5 5 6 8 3 12


Output

2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613

Note

The final positions of the disks will look as follows:

<image>

In particular, note the position of the last disk. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
long double r, rsx[1010], rsy[1010], xx[1010], st, dr, mid, cnt, rr;
long double getdist(long double x, long double y, long double xx,
                    long double yy) {
  return (xx - x) * (xx - x) + (yy - y) * (yy - y);
}
bool check(long double val, int p) {
  for (int i = 1; i <= cnt; i++)
    if (getdist(rsx[i], rsy[i], xx[p], val) - rr <= 0.000000000000001) return 0;
  return 1;
}
int main() {
  cin >> n >> r;
  for (int i = 1; i <= n; i++) cin >> xx[i];
  rsx[1] = xx[1];
  rsy[1] = r;
  rr = 4 * r * r;
  cnt = 1;
  for (int i = 2; i <= n; i++) {
    long double bst = r;
    for (int j = 1; j <= cnt; j++)
      if (abs(xx[i] - rsx[j]) <= 2 * r)
        if (rsy[j] > bst) bst = rsy[j];
    st = bst;
    dr = 4e6;
    for (int j = 1; j <= 80; j++) {
      mid = (st + dr) * 0.5;
      if (check(mid, i))
        dr = mid;
      else
        st = mid;
    }
    rsx[i] = xx[i];
    rsy[i] = (st + dr) * 0.5;
    cnt++;
  }
  for (int i = 1; i <= n; i++)
    cout << fixed << setprecision(10) << rsy[i] << ' ';
  return 0;
}

```