Every move transfers one candy, so the total remains $nm$.

Sort the final counts as $c_1<c_2<\cdots<c_n$. They are nonnegative integers, so $c_i\ge i-1$.

Their total is therefore at least

$0+1+\cdots+(n-1)=\frac{n(n-1)}2.$

If $nm$ is smaller than this, the goal is impossible. Equivalently, output $-1$ when $2m<n-1$.

For the remaining pages, assume $2m\ge n-1$ and write $k=\lfloor n/2\rfloor$. This condition guarantees $m\ge k$. We will show that the goal is possible and find its minimum cost.