$|x - a_1||x - a_2|...|x - a_{2n}| = (n!)^2$

Question

Let $a_i \in \mathbb N$ ; $a_i \neq a_j$ and $i , j = 1 , 2 ,... , 2n$

If $|x - a_1||x - a_2|...|x - a_{2n}| = (n!)^2$. Find $x$ in terms of $a$.

I’ve remove 2n absolute to 1 absolute value of 2n deg polynomial with all as its root.