Hodge Laplacian and scalar

I’m reading Nakahara GEOMETRY, TOPOLOGY AND PHYSICS now.Then, Hodge Laplacian is given by
begin{align}
Delta=(d+d^{dagger})^2=dd^{dagger}+d^{dagger}d
end{align}
For example, we consider 0-form $f$, then $d^{dagger}f=0$
begin{align}
Delta f&=d^{dagger}df=d^{dagger}(partial_{mu}f dx^{mu})
&=-ast d ast(partial_{mu}f dx^{mu})=-ast d(frac{sqrt{|g|}}{(m-1)!}partial_{mu}f g^{mu lambda}epsilon_{lambda nu_2 cdot nu_m} dx^{nu_2}wedge cdots wedge dx^{nu_m})
&=-ast frac{1}{(m-1)!}partial_{nu}[sqrt{|g|}g^{mu lambda} partial_{mu}f] epsilon_{lambda nu_2 cdot nu_m} dx^{nu}wedge dx^{nu_2}wedge cdots wedge dx^{nu_m}
end{align}
So far, so good, but I don’t understand the next transformation.
begin{align}
Delta f&=-ast partial_{nu}[sqrt{|g|}g^{mu lambda} partial_{mu}f] g^{-1} dx^1wedge cdots wedge dx^m
&=-frac{1}{sqrt{|g|}}partial_{nu}[sqrt{|g|}g^{mu lambda} partial_{mu}f]
end{align}
Why did $g^{-1}$ come up? I would appreciate it if you could tell me. For your information, $ast$ is defined as follows.
begin{align}
ast (dx^{mu_1}wedge dx^{mu_2}wedge cdots wedge dx^{mu_r})=frac{sqrt{|g|}}{(m-r)!}epsilon^{mu_1mu_2 cdots mu_r} _{nu_{r+1}cdots nu_{m}}dx^{nu_{r+1}}wedge cdots wedge dx^{nu_{m}}
end{align}
where $m$ is just dimension of manifold.