algorithms – Reduction between Parity-SAT and approximate counting

Consider two problems as defined here.

Approximate counting: Given a Boolean function $f(x)$, for $x in {0, 1}^{n}$, distinguish between the two cases:

  1. The number of satisfying assignments for $f(x)$ is $geq 2^{k+1}$.
  2. The number of satisfying assignments for $f(x)$ is $leq 2^{k}$.

Parity-SAT: Given a Boolean function $f(x)$, for $x in {0, 1}^{n}$, output $1$ if the number of satisfying assignments to $f(x)$ is even.

Is there a way to reduce Parity-SAT to approximate counting (or vice versa)?