2. Let the set $M=\{1,99,-1,0,25,-36,-91,19,-2,11\}$, and denote all non-empty subsets of $M$ as $M_{i}, i=1,2, \cdots$, 2013. The product of all elements in each $M_{i}$ is $m_{i}$. Then $\sum_{i=1}^{2013} m_{i}=$ $\qquad$ .
$$
\sum_{i=1}^{2013} m_{i}=(1+1)(1+99)(1-1)(1+0) \cdots(1+11)-1=-1 \text {. }
$$

Solve:
$$
\sum_{i=1}^{2013} m_{i}=(1+1)(1+99)(1-1)(1+0) \cdots(1+11)-1=-1 \text {. }
$$