Let A be the 4-dimensional algebra over Q with basis 1,i,j,k, and multiplication table $$ i^2 = -1 \quad j^2 = -11 \quad k^2 = -11 \quad ij = k \quad jk = 11i \quad ki = j $$

So, A is the unique "definite quaternion algebra of discriminant 11" over the rational numbers. I'm led to believe that there are two conjugacy classes of maximal orders in A, but I had a hard time finding them explicitly.

A has elements of multiplicative order 4 (e.g. $i$) and $6$, (e.g. $1/2 + i/4 + j/4$). I presume then that $(1,i)$ can be extended to a basis for a maximal order, and that $(1,1/2 + i/4 + j/4)$ can be extended to a basis for a non-conjugate maximal order. What are these bases?

I'd appreciate an answer even just for discriminant 11. Of course any information about how to answer this kind of question systematically would be a nice bonus for me.

PS. Thanks Aeryk and Aurel. One of the two orders, say $O_1$, has basis $$1, \quad i,\quad 1/2+j/2, \quad i/2 + k/2$$ Can anyone tell me the other one? Are the Ivanyos/Ronyai/Voight algorithms good for finding all maximal orders, or would they output something conjugate to $O_1$ again?