Consider $\Bbb{R}^2$ with the usual symplectic form $$\omega = dx \wedge dy$$

For a function $H \colon \Bbb{R}^2 \to \Bbb{R}$, let $X_H$ be the Hamiltonian vector field. Then the map $\Bbb{R}^2 \to \Bbb{R}^2$, defined as the flow of $X_H$ for time $t=1$, is symplectic, and Liouville integrable (since $H$ is invariant).

My question is this: given two functions $H_1(x,y)$ and $H_2(x,y)$, consider the composition of performing the time-1 flow of $X_{H_1}$ followed by the time-1 flow of $X_{H_2}$. Is this map Liouville integrable (i.e. does it have a conserved quantity)?

If so, why? Can the conserved quantity be described in terms of $H_1$ and $H_2$?

If not, is there an easy counter-example?

**Note**: I asked this same quesiton on math stack exchange (here), but got no answers.