Homotopy type of $\mathbb{R}^3 \setminus \{ \mathrm{wedge \hspace{3pt} sum \hspace{3pt} of \hspace{3pt} 2 \hspace{3pt} circles} \}$

Is there something “nice” that the space $ \mathbb{R}^3 \setminus \{ \mathrm{wedge \hspace{3pt} sum \hspace{3pt} of \hspace{3pt} 2 \hspace{3pt} circles} \}$ is homotopy equivalent to? I know that $ \mathbb{R}^3 \setminus \{\mathrm{circle}\} \simeq S^{2} \vee S^{1}$ , and that $ \mathbb{R}^3 \setminus \{ \mathrm{2 \hspace{3pt} linked \hspace{3pt} circles}\} \simeq S^{2} \vee T^{2}$ . However, I can’t seem to apply either of the “visual methods” used to prove these homotopy equivalences to my case.

My wild guess is that it might be homotopic to $ M_{2}$ , the orientable closed surface of genus $ 2$ , but I’m not sure if this is true at all.