#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    long long n, i, j;
    cin >> n;
    double r;
    cin >> r;
    double a[n];
    vector<double> v(n);
    vector<pair<double, long long>> vv;
    cin >> a[0];
    for (i = 1; i < n; i++) {
      cin >> a[i];
      vv.push_back({a[i], i});
    }
    sort(v.begin(), v.end());
    v[0] = r;
    for (i = 1; i < n; i++) {
      long long id = -1;
      double mx = 0;
      for (j = i - 1; j >= 0; j--) {
        if (abs(a[j] - a[i]) <= 2 * r) {
          id = j;
          double y = double(abs(a[i] - a[j])) / ((2.00) * r);
          double z = sqrt(1.00 + y) * sqrt(1.00 - y);
          if (mx < v[j] + (2.00 * r * z)) {
            mx = v[j] + (2.00 * r * z);
          }
        }
      }
      if (id != -1) {
        v[i] = mx;
      } else
        v[i] = r;
    }
    for (i = 0; i < n; i++) printf("%0.12f ", v[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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    long long n, i, j;
    cin >> n;
    double r;
    cin >> r;
    double a[n];
    vector<double> v(n);
    vector<pair<double, long long>> vv;
    cin >> a[0];
    for (i = 1; i < n; i++) {
      cin >> a[i];
      vv.push_back({a[i], i});
    }
    sort(v.begin(), v.end());
    v[0] = r;
    for (i = 1; i < n; i++) {
      long long id = -1;
      double mx = 0;
      for (j = i - 1; j >= 0; j--) {
        if (abs(a[j] - a[i]) <= 2 * r) {
          id = j;
          double y = double(abs(a[i] - a[j])) / ((2.00) * r);
          double z = sqrt(1.00 + y) * sqrt(1.00 - y);
          if (mx < v[j] + (2.00 * r * z)) {
            mx = v[j] + (2.00 * r * z);
          }
        }
      }
      if (id != -1) {
        v[i] = mx;
      } else
        v[i] = r;
    }
    for (i = 0; i < n; i++) printf("%0.12f ", v[i]);
  }
  return 0;
}

```