Let $$q = a_1 x_1^2 + \cdots + a_n x_n^2$$ be a quadratic form over some $p$-adic field $\mathbb{Q}_p$. We thus have its Hasse invariant $$\mathcal{h}(q) = \prod_{1 \leq i < j \leq n} (a_i,a_j)_p \in \{\pm 1\},$$ where $(a_i,a_j)_p$ is the usual Hilbert symbol.
Let $\mathcal{C}(q)$ be the Clifford algebra associated to $q$. If $n$ is even, this is a central simple algebra; in fact, it is a tensor product of quaternion algebras. It thus gives a $2$-torsion element in the Brauer group $\mathbb{Q}/\mathbb{Z}$ of $\mathbb{Q}_p$, i.e. an element $\mathcal{h}'(q) \in \{\pm 1\}$.
Question: How is $\mathcal{h}'(q)$ related to $\mathcal{h}(q)$? We should be able to express $\mathcal{h}'(q)$ in terms of $\mathcal{h}(q)$ along with $n$ and the discriminant, but I don't quite understand all the constructions well enough to do this.