Building on the paper Colin McQuillan suggested, it is easy to see that every $\mathrm{SU}(N)$ matrix $m$ can be decomposed as
$$
m = a \,u\, b
$$
with $a,b \in \mathrm{SU}(N)$ and
$$
u = F \,\left(\frac{1}{\mathrm{det}(V)} \oplus V\right) F^\dagger \in \mathrm{SU}(N)
$$
where $F_{jk} = \frac{1}{\sqrt{N}}e^{\frac{2\pi i}{N}jk}$ is the Fourier transform unitary matrix and $V\in \mathrm{U}(N-1)$.
Notice that the number of parameters of this factorization matches the dimension of $\mathrm{SU}(N)$ .