8. (5 points) In the cells of the table $(2 m-1) \times(2 n-1)$, where $m \geqslant n$, the numbers 1, 2, and 3 are arranged such that in any $2 \times 2$ square there are all three different numbers. What is the smallest value that the sum of the numbers in the entire table can take?
Answer: $7 m n-5 m-4 n+3$.