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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.

Solution
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