Is there an invariant for mapping $(p,q) \mapsto (\frac{p}{1-(1-p)q}, \frac{qp}{1-(1-p)q})$?

Question

Suppose starting with $0<p,q<1$, mapping $f(p,q)$ that maps $(p,q)$ to $f(p,q) = \left(\frac{p}{1-(1-p)q}, \frac{qp}{1-(1-p)q}\right)$.

Is there an invariant for this mapping $f$, such as $g$, so that $g(p,q) = g(f(p,q))$?