A sequence $\{x_n \}=x_0, x_1, x_2, \cdots $ satisfies $x_0=a(1\le a \le 2019, a \in \mathbb{R})$, and $$x_{n+1}=\begin{cases}1+1009x_n &\ (x_n \le 2) \\ 2021-x_n &\ (2<x_n \le 1010) \\ 3031-2x_n &\ (1010<x\le 1011) \\ 2020-x_n &\ (1011<x_n) \end{cases}$$for each non-negative integer $n$. If there exist some integer $k>1$ such that $x_k=a$, call such minimum $k$ a fundamental period of $\{x_n \}$. Find all integers which can be a fundamental period of some seqeunce; and for such minimal odd period $k(>1)$, find all values of $x_0=a$ such that the fundamental period of $\{x_n \}$ equals $k$.