In this paper https://arxiv.org/abs/hep-th/9906127 (see eq. 15)
The following identity appears
$$ \delta_{W} \int d^d x \sqrt{-\gamma} \tilde{\mathcal{L}}^{(n)} = \int d^d x \sqrt{-\gamma} \sigma\left( \frac{d-2n}{2} \tilde{\mathcal{L}}^{(n)} + \nabla_{a} X^{(n)a} \right) $$
where
$$\delta_{W} \gamma_{ab} = \sigma \gamma_{ab}$$
I am trying to derive the above identity.
$$ \delta_{W} (\sqrt{-\gamma}\mathcal{L}^{(n)}) = \mathcal{L}^{(n)}\delta_{W} (\sqrt{-\gamma}) + \sqrt{-\gamma}\delta_{W} (\mathcal{L}^{(n)}) $$
Now
$$\delta_{W}\sqrt{-\gamma} = \frac{1}{2} \sqrt{-\gamma} \gamma^{ab}\delta_{W}\gamma_{ab} = \frac{\sigma d}{2}\sqrt{-\gamma}$$
Now how do I show that
$$\sqrt{-\gamma}\delta_{W} (\mathcal{L}^{(n)}) = -n \sqrt{-\gamma} \sigma \mathcal{L}^{(n)} + \nabla_{a}X^{(n)a}$$
where $X^{(n)a}$ is unspecified and depends on the form of $\mathcal{L}^{(n)}$
