Matrix Sum, factorial(K) times
Suppose that the original elements are $a_{ij}^{(1)}$. Their sum is $S^{(1)} = \sum_{ij} a_{ij}^{(1)}$, and the new elements are $a_{ij}^{(2)} = a_{ij}^{(1)} + S^{(1)}$. The new sum is $S^{(2)} = \sum_{ij} a_{ij}^{(1)} + nmS^{(1)} = (1+nm)S^{(1)}$, and the new elements are $a_{ij}^{(3)} = a_{ij}^{(1)} + (2+nm)S^{(1)}$. The new sum is $S^{(3)} = S^{(1)} + nm(2+nm)S^{(1)} = (1+nm)^2S^{(1)}$, and so on.

More generally, you can prove by induction that $S^{(t)} = (1+nm)^{t-1} S^{(1)}$ and $a_{ij}^{(t)} = a_{ij}^{(1)} + \frac{(1+nm)^{t-1}-1}{nm} S^{(1)}$. It remains to find a fast way to calculate expressions such as $S^{(k!)} = (1+nm)^{k!-1} S^{(1)}$ modulo the given prime $p = 10^9 + 7$. You do the rest.