AI Finds Nothing Here

Think of every cell as a variable over $\mathbb F_2$, because XOR is just addition modulo $2$. The condition for each $r \times c$ submatrix is one linear equation:

$$\sum_{x=i}^{i+r-1}\sum_{y=j}^{j+c-1} a_{x,y}=0 \pmod 2.$$

So the number of valid matrices is $2^d$, where $d$ is the number of free variables after all independent constraints are applied.

The only real job is counting how many of those constraints are independent. Brute force is dead on arrival since $n,m$ go up to $10^9$, so we need the structure.

Define a 1D interval-sum matrix $R$ for rows. It has $n-r+1$ rows and $n$ columns, where row $i$ has ones in columns $i,i+1,\dots,i+r-1$. Similarly define $C$ for columns, with size $(m-c+1) \times m$.

For a matrix $A$, all scanned XORs together are exactly:

$$RAC^T.$$

We need:

$$RAC^T=0.$$

Now count the rank of this linear map.

First, the 1D interval-sum matrix has full row rank. For $R$, the first row has a $1$ in column $1$, and no later row has a $1$ there. So row $1$ cannot be made from later rows. Remove it and repeat the same argument on column $2$, then column $3$, and so on. More formally, each row has a unique leftmost column among the remaining rows, so all rows are independent.

Therefore:

$$\operatorname{rank}(R)=n-r+1,$$

and similarly:

$$\operatorname{rank}(C)=m-c+1.$$

The map $A \mapsto RAC^T$ has rank

$$\operatorname{rank}(R)\operatorname{rank}(C).$$

One way to see this: by invertible row/column changes, $R$ can be transformed into a matrix that keeps exactly $\operatorname{rank}(R)$ independent row coordinates, and $C$ into one that keeps exactly $\operatorname{rank}(C)$ independent column coordinates. Then $RAC^T$ simply exposes a block of size

$$\operatorname{rank}(R) \times \operatorname{rank}(C),$$

so exactly that many independent linear constraints exist. No hidden extra constraints, no magic, no AI finding anything here. Just rank.

Thus the number of free variables is:

$$d = nm - (n-r+1)(m-c+1).$$

Every assignment to those $d$ free variables determines one valid matrix, so the answer is:

$$2^d \bmod 998244353.$$

Edge cases also fall out cleanly:

• If $r=c=1$, every single cell must be $0$, so $d=nm-nm=0$ and the answer is $1$.
• If $r=n$ and $c=m$, there is only one constraint: the XOR of the whole matrix is $0$, so $d=nm-1$.
• If only one scan dimension is $1$, the formula still handles it because the corresponding interval matrix has full row rank.

The exponent can be around $10^{18}$, so use long long. Fast exponentiation gives each test case in $O(\log d)$ time.

Complexity per test case:

$$O(\log(nm))$$

with $O(1)$ memory.