# Does Barvinok's algorithm count modulo $q$ in $O(polylog (q))$ word size?

Let $Ax\leq b$ be a polyhedron in $n$-dimensions and $m$ constraints and $q>1$ be an integer. The number of points in the polyhedron could be exponential in $n$ and $m$ while $q\ll nm$ could be true. If we have to count number of points and then take modulo $q$ our word size will be $poly(nm)$ (since $O(nm)$ bits are needed to represent the number of points as an integer). So can we count modulo $q$ in $O(polylog (q))$ word size which is $O(polylog (nm))$ bits using Barvinok's algorithm?