Solve the following math problem efficiently and clearly:

- For simple problems (2 steps or fewer):
Provide a concise solution with minimal explanation.

- For complex problems (3 steps or more):
Use this step-by-step format:

## Step 1: [Concise description]
[Brief explanation and calculations]

## Step 2: [Concise description]
[Brief explanation and calculations]

...

Regardless of the approach, always conclude with:

Therefore, the final answer is: $oxed{answer}$. I hope it is correct.

Where [answer] is just the final number or expression that solves the problem.

Problem: A $\frac 1p$ -array is a structured, infinite, collection of numbers. For example, a $\frac 13$ -array is constructed as follows:
\begin{align*} 1 \qquad \frac 13\,\ \qquad \frac 19\,\ \qquad \frac 1{27} \qquad &\cdots\\ \frac 16 \qquad \frac 1{18}\,\ \qquad \frac{1}{54} \qquad &\cdots\\ \frac 1{36} \qquad \frac 1{108} \qquad &\cdots\\ \frac 1{216} \qquad &\cdots\\ &\ddots \end{align*}
In general, the first entry of each row is $\frac{1}{2p}$ times the first entry of the previous row. Then, each succeeding term in a row is $\frac 1p$ times the previous term in the same row. If the sum of all the terms in a $\frac{1}{2008}$ -array can be written in the form $\frac mn$, where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by $2008$.
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