A linearly ordered (shortly, l.o.) monoid is a triple $\mathbb M = (M, \cdot, \le)$ for which $(M, \cdot)$ is a (multiplicatively written) monoid and $\le$ is a total order on $M$ such that $xy < xz$ and $yx < zx$ for all $x,y,z \in M$ with $y < z$. In particular, $\mathbb M$ is called an l.o. group if $(M, \cdot)$ is, well, a group, and an l.o. free monoid if $(M, \cdot)$ is the free monoid on an alphabet $X$.

What is known about the following question?

(Q) Let $\mathbb M = (M, \cdot, \le)$ be a linearly ordered free monoid. Does there always exist an embedding of $\mathbb M$ into a linearly ordered group? In more plain words: do there always exist a linearly ordered group $\mathbb G = (G, \cdot, \le)$ and a (monoid) monomorphism $f: (M, \cdot) \to (G, \cdot)$ such that $f(x) < f(y)$ for all $x,y \in M$ with $x < y$?

The answer is affirmative in the case when $\le$ is the lexicographic order induced on $M$ by any well-ordering of the underlying alphabet (this can be proved, e.g., by the "Magnus trick").

But what about the rest? Any reference?