Determining the Jordan Canonical Form $18\times 18$ matrix

Determining the Jordan Canonical Form $18\times 18$ matrix

Let A be an $18\times 18$ matrix over $\mathbb{C}$ with characteristic
polynomial equal to $$ (x-1)^6(x-2)^6(x-3)^6 $$ and a minimal polynomial
equal to $$ (x-1)^4(x-2)^4(x-3)^3. $$ Assume $(A-I)$ has nullity 2,
$(A-2I)$ has nullity 3, and $(A-3I)^2$ has nullity 4. Find the Jordan
canonical form of A.
For $\lambda=1$, I figure the largest block is a $4\times 4$ (since the
multiplicity of the root is 4 in the minimal polynomial), and since the
nullity is 2, that there are 2 blocks in total, thus forcing the other
block to be 2x2.
For $\lambda=2$, I figure the largest block is a 4x4, there are 3 blocks,
thus making the others $1\times 1$ matrices.
For $\lambda=3$, The largest block is a $3\times 3$ matrix. However, I am
not sure how the nullity of $(A-3I)^2$ determines the nullity of $(A-3I)$.
Any insight into this problem would be extremely valuable as I am trying
to teach myself this process.