Covering $0$s and $1$s of a matrix

Question

Let $M \in \{0,1\}^{n \times n}$ be a binary matrix. A combinatorial rectangle $R \subseteq [n] \times [n]$ is a set such that there exists $X \subseteq [n]$ and $Y \subseteq [n]$ where $R = X \times Y$. A $0$-rectangle is a rectangle $R$ such that for all $(i,j) \in R$, $M_{i,j} = 0$. Suppose that there are $t$ $0$-rectagnles $R_1, \ldots, R_t$ covering all the zeros in the matrix, i.e. for all $(i,j)$ such that $M_{i,j} = 0$ there is some rectangle $R_k$ such that $(i,j) \in R_k$. My question is: how many $1$-rectangles do I need in order to cover all the $1$s of $M$? (i.e. in the worst case; think about an adversary trying to force me to use as many rectangles as possible, but this adversary must be able to cover the $0$s with at most $t$ $0$-rectangles). My guess is that $2^t$ $1$-rectangles will be enough, no matter where the $0$s are, but I have to idea how to prove it. Any help will be appriciated.