6) Let $\left\{a_{n}\right\}$ be a sequence with the sum of the first $n$ terms denoted as $S_{n}$. Let $T_{n}=\frac{S_{1}+S_{2}+\cdots+S_{n}}{n}$, which is called the "average" of the sequence $a_{1}, a_{2}, \cdots, a_{n}$. Given that the "average" of the sequence $a_{1}, a_{2}, \cdots, a_{1005}$ is 2012, what is the "average" of the sequence $-1, a_{1}, a_{2}, \cdots, a_{1005}$? $\qquad$ .
62009 Prompt: Let $A_{n}=S_{1}+S_{2}+\cdots+S_{n}$, then
$$
T_{1005}=\frac{S_{1}+S_{2}+\cdots+S_{1005}}{1005}=\frac{A_{1005}}{1005}=2012 \text {, }
$$

Thus, $A_{1005}=2012 \times 1005$. Therefore,
$$
\begin{aligned}
T_{1006} & =\frac{-1+\left(-1+S_{1}\right)+\left(-1+S_{2}\right)+\cdots+\left(-1+S_{1005}\right)}{1006} \\
& =\frac{-1006+A_{1005}}{1006}=\frac{-1006+2012 \times 1005}{1006} \\
& =-1+2 \times 1005=2009 .
\end{aligned}
$$