Separably rationally connected surface is rational

Question

A variety over a field $k$ is called separably rationally connected if there is a variety Y along with a morphism $u:U=Y \times \mathbb{P}^1 \to X$ such that the map $u \times u : U \times_{Y} U \to X \times X$ is dominant and smooth at generic point. Then prove that separably rationally connected surface is rational.
I think that first we need to show that a separably rationally connected variety is unirational. But I can't get a separable dominant map $\mathbb{P}^2 \to X$.