Taylor expansion of determinant of Riemannian metric in normal coordinates up to higher order

Let $$(M,g)$$ be an $$n$$-dimensional Riemannian manifold. Let $$p\in M$$, and let $$\{x^i\}_{i=1}^n$$ be normal coordinates centered around $$p$$.

Using Jacobi field, one can show that the metric $$g$$ has the following Taylor expansion \begin{align} g_{ij}(x)&=\delta_{ij}-\frac{1}{3}R_{ipqj}(p)x^px^q-\frac{1}{6}\nabla_rR_{ipqj}(p)x^px^qx^r \\ &\qquad+\left(-\frac{1}{20}\nabla_r\nabla_rR_{ipqj}(p)+\frac{2}{45}g^{kl}R_{ipqk}R_{jrsl}(p)x^px^qx^rx^s\right)+O(|x|^5) \end{align} where $$\nabla$$ is the Levi-Civita connection of $$(M,g)$$, $$R_{ijkl}$$ are the (components of the) Riemann curvature tensor, while $$x$$ is a point near $$p$$ with coordinates $$x^i$$, and $$|x|:=d(x,p)$$, the radial distance from $$p$$.

Using this, together with the Jacobi formula for derivative of determinant function, one should be able to obtain the Taylor expansion of $$\det(g_{ij})$$. It is claimed (e.g. in Hamilton's Ricci flow page 59) that \begin{align} \det(g_{ij})(x)&=1-\frac{1}{3}R_{ij}(p)x^ix^j-\frac{1}{6}\nabla_kR_{ij}(p)x^ix^jx^k \\ &\quad-\left(\frac{1}{20}\nabla_l\nabla_kR_{ij}(p)+\frac{1}{90}g^{pq}g^{rs}R_{pijr}R_{qkls}(p)-\frac{1}{18}R_{ij}R_{kl}(p)\right)x^ix^jx^kx^l \\ &\quad+O(|x|^5) \end{align} where $$R_{ij}$$ are the (components of the) Ricci curvature tensor.

My question is that

How do we obtain the term \begin{align} \frac{1}{90}g^{pq}g^{rs}R_{pijr}R_{qkls}(p)-\frac{1}{18}R_{ij}R_{kl}(p) \end{align}

I believe it should come from the term $$\displaystyle\frac{2}{45}g^{kl}R_{ipqk}R_{jrsl}(p)$$ in the expansion of $$g_{ij}$$. By using Jacobi's formula and evaluating at $$p$$ (since $$p$$ is the point where $$x=0$$, many terms will vanish), one should see that the coefficient of $$x^ix^jx^kx^l$$ is just \begin{align} \frac{1}{4!}g^{ab}\partial_i\partial_j\partial_k\partial_lg_{ab}(p) \end{align} where $$\partial_i=\frac{\partial}{\partial x^i}$$. In other word, I expect that we only need to take the trace (and take care of the counting of possibly repeated terms). But then tracing $$\displaystyle\frac{2}{45}g^{kl}R_{ipqk}R_{jrsl}(p)$$ doesn't seem to yield the desired result. In particular, I am not quite sure how $$R_{ij}R_{kl}$$ pops out.

Any comment, hint or answer are greatly appreciated.

$$\partial^4_{ijkl} \det(g) = \partial^3_{ijk} (g^{-1} \partial_l g) = g^{-1} \partial^4_{ijkl} g + ( \partial^2_{ij} g^{-1} \partial^2_{kl} g + \partial^2_{ik} g^{-1} \partial^2_{jl} + \partial^2_{jk} g^{-1} \partial^2_{il} g)$$
$$\partial^2_{ij} g^{-1} = -g^{-1} (\partial^2_{ij} g) g^{-1}$$
after dropping the first order terms. So you should get contributions both from the quartic part of the expansion of $$g$$ and the quadratic part of the expansion of $$g$$, the latter of which you don't seem to have accounted for.