The Sylow Theorems · Challenge problem

A group $G$ has order $546=2\cdot3\cdot7\cdot13$. Suppose, toward a contradiction, that NEITHER its Sylow $7$-subgroup nor its Sylow $13$-subgroup is normal, so Sylow III forces $n_7=78$ and $n_{13}=14$ (the only other candidates for each). How many elements of order $7$ or order $13$ would this force $G$ to have in total?