#include <bits/stdc++.h>
#pragma comment(linker, "/stack:20000000")
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
inline bool isPrime(long long int n) {
  if (n <= 1) {
    return false;
  }
  if (n <= 3) {
    return true;
  }
  if (n % 2 == 0 || n % 3 == 0) {
    return false;
  }
  for (int i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long int n, r;
  cin >> n >> r;
  long long int t[n];
  for (long long int i = 0; i < n; i++) {
    cin >> t[i];
  }
  long double ans[n];
  ans[0] = r;
  for (long long int i = 1; i < n; i++) {
    long double x = 0;
    for (long long int j = 0; j < i; j++) {
      if (abs(t[i] - t[j]) <= 2 * r) {
        long double y = sqrt(4 * r * r - (t[i] - t[j]) * (t[i] - t[j]));
        x = max(x, ans[j] + y);
      }
    }
    ans[i] = max(x, (long double)r);
  }
  for (long long int i = 0; i < n; i++) {
    cout << setprecision(20) << fixed << ans[i] << " ";
  }
  cout << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#pragma comment(linker, "/stack:20000000")
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
inline bool isPrime(long long int n) {
  if (n <= 1) {
    return false;
  }
  if (n <= 3) {
    return true;
  }
  if (n % 2 == 0 || n % 3 == 0) {
    return false;
  }
  for (int i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long int n, r;
  cin >> n >> r;
  long long int t[n];
  for (long long int i = 0; i < n; i++) {
    cin >> t[i];
  }
  long double ans[n];
  ans[0] = r;
  for (long long int i = 1; i < n; i++) {
    long double x = 0;
    for (long long int j = 0; j < i; j++) {
      if (abs(t[i] - t[j]) <= 2 * r) {
        long double y = sqrt(4 * r * r - (t[i] - t[j]) * (t[i] - t[j]));
        x = max(x, ans[j] + y);
      }
    }
    ans[i] = max(x, (long double)r);
  }
  for (long long int i = 0; i < n; i++) {
    cout << setprecision(20) << fixed << ans[i] << " ";
  }
  cout << '\n';
  return 0;
}

```