Non-isomorphic countable linear orders which embed into each other

Question

Def: $(A,<) \hookrightarrow (B,\lhd)$ if $\exists f:A \to B$ such that $\forall a,b \in A[a<b \implies f(a) \lhd f(b)]$.

May I seek your advice on how to construct two infinite, countable linear orders $(A,<)$ and $ (B,\lhd)$ such that $(A,<) \hookrightarrow (B,\lhd)$ and $(B,\lhd) \hookrightarrow (A,<)$, but $(A,<)$ and $(B,\lhd)$ are not isomorphic?

Thank you.

Answer 1

HINT:

Note that $\Bbb Q$ is isomorphic to $\Bbb Q\cap(0,1)$; but not to $\Bbb Q\cap[0,1]$. Both with the standard order.