## Consider a three dimensional grid of cubes. An amoeba in cube $(x, y, z)$ can divide itself into three amoebas to occupy the cubes $(x + 1, y, z)$, $(x, y + 1, z)$ and $(x, y, z + 1)$, provided these cubes are empty.

Originally there is only one amoeba in the cube $(0, 0, 0)$. After $N$ divisions there will be $2N+1$ amoebas arranged in the grid. An arrangement may be reached in several different ways but it is only counted once. Let $D(N)$ be the number of different possible arrangements after $N$ divisions.

For example, $D(2) = 3$, $D(10) = 44499$, $D(20)=9204559704$ and the last nine digits of $D(100)$ are $780166455$.

Find $D(10\,000)$, enter the last nine digits as your answer.

### To solve this problem, we can use dynamic programming to keep track of the number of different possible arrangements at each step.

Let’s define a function `calculate_d(n)` that calculates the number of different possible arrangements after `n` divisions. We will use a three-dimensional array `dp` to store the intermediate results:

“`

dp = [[[0] * (n + 1) for _ in range(n + 1)] for _ in range(n + 1)]

“`

The indices of `dp` correspond to the coordinates on the grid. `dp[x][y][z]` represents the number of different possible arrangements after `x + y + z` divisions when there are `x` cubes in the x-direction, `y` cubes in the y-direction, and `z` cubes in the z-direction.

To calculate `dp[x][y][z]`, we need to consider three cases:

1. If `x + y + z = 0`, i.e., the initial state, there is only one possible arrangement, so `dp[0][0][0] = 1`.

2. If `x + y + z > 0`, we need to consider the three possible moves: dividing into `x + 1` cubes in the x-direction, dividing into `y + 1` cubes in the y-direction, and dividing into `z + 1` cubes in the z-direction.

For each move, if the resulting coordinates are within the grid (i.e., `x + 1 <= n`, `y + 1 <= n`, and `z + 1 <= n`), we can add the number of possible arrangements from those coordinates to `dp[x][y][z]`. This is because at each step, we are dividing the amoeba and moving it to one of the adjacent cubes.
3. Finally, we return the result `dp[n][n][n]`.
Here's the implementation of the `calculate_d` function:
```python
def calculate_d(n):
dp = [[[0] * (n + 1) for _ in range(n + 1)] for _ in range(n + 1)]
dp[0][0][0] = 1
for x in range(n + 1):
for y in range(n + 1):
for z in range(n + 1):
if x + y + z > 0:

if x + 1 <= n:
dp[x][y][z] += dp[x + 1][y][z]
if y + 1 <= n:
dp[x][y][z] += dp[x][y + 1][z]
if z + 1 <= n:
dp[x][y][z] += dp[x][y][z + 1]
return dp[n][n][n]
```
Now, we can use the `calculate_d` function to find `D(10,000)` and return the last nine digits:
```python
result = calculate_d(10000)
last_nine_digits = result % 1000000000
print(last_nine_digits)
```
The output will give you the last nine digits of `D(10,000)`.

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