#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  double r;
  scanf("%d %lf", &n, &r);
  double x[1005];
  double ansy[1005];
  for (int i = 1; i <= n; i++) {
    scanf("%lf", &x[i]);
  }
  ansy[1] = r;
  for (int i = 2; i <= n; i++) {
    double sx = x[i] - r;
    double sy = x[i] + r;
    int maxi = -1;
    ansy[i] = r;
    for (int j = 1; j < i; j++) {
      double sxx = x[j] - r;
      double syy = x[j] + r;
      if ((sxx >= sx && syy <= sy) || (sxx <= sx && syy <= sy && syy >= sx) ||
          (sxx <= sy && sxx >= sx && syy >= sy)) {
        maxi = j;
        double rr = r * r;
        double diffx = (x[maxi] + (double)x[i]) / 2.0 - x[maxi];
        diffx = diffx * diffx;
        double left = sqrt(rr - diffx);
        double ycord = left * 1.00 + ansy[maxi];
        diffx = (x[maxi] * 1.00 + (double)x[i]) / 2.0 - x[i] * 1.00;
        diffx = diffx * diffx;
        left = (double)sqrt(rr * 1.00 - diffx * 1.00);
        ansy[i] = max(ycord + left * 1.00, ansy[i]);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%0.10lf\n", ansy[i]);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 main() {
  int n;
  double r;
  scanf("%d %lf", &n, &r);
  double x[1005];
  double ansy[1005];
  for (int i = 1; i <= n; i++) {
    scanf("%lf", &x[i]);
  }
  ansy[1] = r;
  for (int i = 2; i <= n; i++) {
    double sx = x[i] - r;
    double sy = x[i] + r;
    int maxi = -1;
    ansy[i] = r;
    for (int j = 1; j < i; j++) {
      double sxx = x[j] - r;
      double syy = x[j] + r;
      if ((sxx >= sx && syy <= sy) || (sxx <= sx && syy <= sy && syy >= sx) ||
          (sxx <= sy && sxx >= sx && syy >= sy)) {
        maxi = j;
        double rr = r * r;
        double diffx = (x[maxi] + (double)x[i]) / 2.0 - x[maxi];
        diffx = diffx * diffx;
        double left = sqrt(rr - diffx);
        double ycord = left * 1.00 + ansy[maxi];
        diffx = (x[maxi] * 1.00 + (double)x[i]) / 2.0 - x[i] * 1.00;
        diffx = diffx * diffx;
        left = (double)sqrt(rr * 1.00 - diffx * 1.00);
        ansy[i] = max(ycord + left * 1.00, ansy[i]);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%0.10lf\n", ansy[i]);
  }
  return 0;
}

```