Consider two sets $A$ and $B$, each containing three numbers in A.P. Let the sum and the product of the elements of $A$ be $36$ and $p$ respectively and the sum and the product of the elements of $B$ be $36$ and $q$ respectively. Let $d$ and $D$ be the common differences of the A.P.s in $A$ and $B$ respectively such that $D=d+3$, $d>0$. If $\dfrac{p+q}{p-q}=\dfrac{19}{5}$, then $p-q$ is equal to: