Suppose $p,q$ are distinct primes with least quadratic non-residues $n_p$ and $n_q$ respectively. Can one bound the least $n$ for which $\left(\frac{n}{p}\right)=\left(\frac{n}{q}\right)=-1$ in terms of $n_p$ and $n_q$?

I had originally thought this would be a consequence of quadratic reciprocity and the Chinese Remainder Theorem, and so would be bounded by $n_pn_q$, but I can't seem to work it out.