The Sylow Theorems · Superchallenge problem

A group has order $380=2^2\cdot5\cdot19$. Suppose, toward a contradiction, that NEITHER its Sylow $5$-subgroup nor its Sylow $19$-subgroup is normal, so Sylow III forces $n_5=76$ and $n_{19}=20$ (the only other candidates). How many elements of order $5$ or order $19$ would this force $G$ to have in total?