Coherent trees that are not Countryman

First, we show that every coherent tree that contains a Countryman suborder is ({mathbb {R}})-embeddable when restricted to a club. Then for a linear order O that can not be embedded into (omega ), there exists (consistently) an ({{mathbb {R}}})-embeddable O-ranging coherent tree which is not Countryman. And for a linear order (O') that can not be embedded into ({mathbb {Z}}), there exists (consistently) an ({mathbb {R}})-embeddable (O')-ranging coherent tree which contains no Countryman suborder. Finally, we will see that this is the best we can do.