Parity of a fermion bilinear

I’m assuming that the parity transformation of a 4-vector field is:

$$x^mu = (t,mathbf{x}) rightarrow x’^mu = (t’,mathbf{x’}) = (t,-mathbf{x})$$
$$V^mu(t,mathbf{x}) = (V^0(t,mathbf{x}), V^i(t,mathbf{x})) rightarrow V’^mu(t’,mathbf{x’}) = (V^0(t,mathbf{x}), – V^i(t,mathbf{x}))$$

The parity transformation of a Dirac spinor is:

$$ Psi (t,mathbf{x}) rightarrow Psi’ (t’,mathbf{x’}) = gamma^0 Psi (t’,mathbf{x’}) = gamma^0 Psi (t,-mathbf{x}) $$

Then de bilinear $j^mu_V (t,mathbf{x})= bar{Psi} (t,mathbf{x}) gamma^mu Psi (t,mathbf{x})$ must transform under parity as:

$$j^mu_V (t,mathbf{x}) rightarrow j’^mu_V (t’,mathbf{x’}) = bar{Psi}'(t’,mathbf{x’}) gamma^mu Psi’ (t’,mathbf{x}’) = bar{Psi} (t,-mathbf{x}) gamma^0 gamma^mu gamma^0 Psi (t,-mathbf{x}) = (j^0_V (t,-mathbf{x}),-j^i_V (t,-mathbf{x}))$$

Where I have used the relation $ { gamma^mu,gamma^nu } = 2 eta^{mu nu} $. This result is almost the parity transformation of the 4-vector field above, the $(t,-mathbf{x})$
dependency should be $(t,mathbf{x})$ for a perfect match. Then, why is this current considered a polar vector if it doesn’t transform as such? What am I getting wrong?