#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkmax(T& a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
bool chkmin(T& a, T b) {
  return a > b ? a = b, 1 : 0;
}
const int maxn = 1000 + 10;
int n;
double r;
double x[maxn], y[maxn];
int main() {
  scanf("%d", &n);
  scanf("%lf", &r);
  for (int i = 1; i <= n; ++i) scanf("%lf", &x[i]);
  y[1] = r;
  for (int i = 2; i <= n; ++i) {
    int ok = 1;
    for (int j = 1; j <= i - 1; ++j) {
      double dis = (x[i] > x[j] ? x[i] - x[j] : x[j] - x[i]);
      double t = 2 * r;
      if (dis > t && fabs(dis - t) >= 0.000000001) continue;
      if (chkmax(y[i], y[j] + sqrt(t * t - dis * dis))) ok = 0;
    }
    if (ok) y[i] = r;
  }
  for (int i = 1; i <= n; ++i) printf("%.10lf ", y[i]);
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T>
bool chkmax(T& a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
bool chkmin(T& a, T b) {
  return a > b ? a = b, 1 : 0;
}
const int maxn = 1000 + 10;
int n;
double r;
double x[maxn], y[maxn];
int main() {
  scanf("%d", &n);
  scanf("%lf", &r);
  for (int i = 1; i <= n; ++i) scanf("%lf", &x[i]);
  y[1] = r;
  for (int i = 2; i <= n; ++i) {
    int ok = 1;
    for (int j = 1; j <= i - 1; ++j) {
      double dis = (x[i] > x[j] ? x[i] - x[j] : x[j] - x[i]);
      double t = 2 * r;
      if (dis > t && fabs(dis - t) >= 0.000000001) continue;
      if (chkmax(y[i], y[j] + sqrt(t * t - dis * dis))) ok = 0;
    }
    if (ok) y[i] = r;
  }
  for (int i = 1; i <= n; ++i) printf("%.10lf ", y[i]);
  return 0;
}

```