Is there a neat way to construct the Coxeter Group $H_4$ from that of $H_3$ using the fact that $|H_4| = 14400 = 120^2 =|H_3|^2$?

Question

Let $H_4$ and $H_3$ be the usual finite irreducible Coxeter Groups of their names: as presentations $H_4 = \langle s_1,s_2,s_3,s_4\rangle$ subject to the relations $$s_i^2 = (s_is_j)^2 = (s_1s_2)^5 = (s_2s_3)^3 = (s_3s_4)^3$$ for $i,j \in \{1,2,3,4\}$ and $|i-j|\ge 2$ and $H_3 = \langle s_1,s_2,s_3\rangle$ subject to those same relations that contain only elements that are $s_1,s_2$ or $s_3$. These both happen to be finite groups, where $$|H_4| = 14400 = 120^2 =|H_3|^2.$$ What are some nice ways one might construct $H_4$ from $H_3$ using the the fact that one has order that is square of the other?