A question about $\delta(x) \star \delta(p)$

Using the Moyal product between two delta functions in $$(x,p)$$-space one gets $$\delta(x) \star \delta(p) = \frac{1}{\pi} e^{2ixp}.$$

However, $$\delta(-x)=\delta(x)$$ and last time I checked $$e^{2ixp} \neq e^{-2ixp}$$.

What's going on here?

Recall the star operator also depends on x, $$\delta (x) ~ \star ~ \delta(p) = \delta (x) ~ \exp{\left( \frac{i \hbar}{2} \left(\overleftarrow {\partial }_x \overrightarrow{\partial }_p-\overleftarrow{\partial }_p \overrightarrow {\partial }_x \right) \right)} ~ \delta(p) ={2\over h} \exp \left (2i{xp\over\hbar}\right ).$$ Consequently, flipping the sign of x complex conjugates it, even though it leaves the delta function invariant.