Modeling equality in an ILP

Question

Lets say we have integer variables $a \in\mathbb{Z}^n$ and $M \in\{0,1\}^{n\times L}$. I am promised $a_i \leq L$, for some fixed constant $L$. I want to model the constraint $$M_{i,j} \iff (a_i=j)$$ in an integer linear program. What is the most efficient way to do this in an ILP? In particular, can I do this without Big-M constraints and without breaking down each integer into its binary representation?

Currently I can see two ways to model this:

Option 1

We can apply the method described in the answer here to the constraint $M_{i,j} \iff (a_i-j=0)$. This requires $1$ extra binary variable as well as $4$ potentially large "Big M" constraints per $M_{i,j}$, thus resulting in $nL$ new binary variables and $4nL$ new big M constraints.

Option 2

We can break down $a_i$ into its bits using $\log L$ extra binary variables. We can also compute the corresponding binary bit breakdowns for each $0\leq j \leq L$. Let $bit^{(a)}_{i,k}$ be the $k$th bit of $a_i$ and let $bit_{j,k}^*$ be the $k$th bit of the integer $j$, so that $$a_i = \sum_{k=0}^{\log L}2^kbit^{(a)}_{i,k}.$$

We now have the following logic: $$M_{i,j} \iff (\forall k . 0\leq k\leq \log L \implies bit^{(a)}_{i,k} \odot bit_{j,k}^*)$$

Therefore

$$M_{i,j} = \bigwedge_{k=0}^{\log L} bit^{(a)}_{i,k} \odot bit_{j,k}^*.$$

We can use this post to NXOR the pairs of variables, storing their results as another variable $z=\{0,1\}^{\log L}$. We can AND these together using the constraints $$M_{i,j} \ge \sum z_i$$ $$M_{i,j} \leq z_i \;\;\; \forall i$$

This creates $\log L$ new variables and 1 constraint for representing the binary breakdown of $a_i$. We then have $\log L$ XNOR variables, each which can be represented with 4 constraints. We can finally represent $M_{i,j}$ with $\log L + 1$ more constraints. Therefore, we have a total of $5\log L$ constraints and $2\log L$ new variables per $M_{i,j}$, giving us a total of $5nL\log L$ new binary constraints and $2nL\log L$ new binary variables.

Is there some third option that is better than either of these two options?