(2) Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=2, a_{2}=1, a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+$ $a_{n+2}\left(n \in \mathbf{N}^{*}\right)$, then $a_{1}+a_{2}+\cdots+a_{2011}=$ $\qquad$ .