Do all finitely generated nilpotent semigroups have polynomial growth?

The notion of nilpotency passes nicely from groups to semigroups. Define $q_1(x,y)=xy$ and $$q_{i+1}(x,y,z_1,\cdots,z_i)=q_i(x,y,z_1,\cdots,z_{i-1})z_iq_i(y,x,z_1,\cdots,z_{i-1})$$ inductively for all $x,y,z_1,z_2,\cdots$.

A. I. Malcev proved that nilpotent groups of class $c$ are described as the groups satisfying the law $q_c(x,y,z_1,\ldots,z_{c-1})=q_c(y,x,z_1,\ldots,z_{c-1})$. It is natural to define nilpotent semigroups of class $c$ as those satisfying this law.

Question: Do all f.g. nilpotent semigroups have polynomial growth?

Note that for cancellative semigroups the answer is "yes" since nilpotent cancellative semigroups satisfy the Ore's condition and so are group-embeddable.

(My guess is that in general the answer is "no" and there even perhaps exists a counter-example among matrix semigroups.)

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The answer is "no", see Theorem 6.1 in Shneerson, L. M. Relatively free semigroups of intermediate growth. J. Algebra 235 (2001), no. 2, 484–546. The relatively free semigroup in the variety $xyzyx=yxzxy$ ("nilpotent" of class 2) with at least 3 generators has intermediate growth ($\sim 2^{\sqrt{n}}$).