Suppose we are considering $S_n$. For any permutation, let $h$ be the number of derangement and $N$ be the number of cycles with length no less than 2.

I'm interested in the number of permutations such that $h-N=k$. I do not need an exact value. An upper bound depending on $n$ and $k$ for large $n$ is enough. I also have a guess of the order of the upper bound: $$(n/\sqrt{2k})^{2k}$$ I come up with this guess due to the fact that if $h-N=k$, $h$ must be smaller or equal to $2k$. In the case $h=2k$, we have $$\frac{n(n-1)\cdots(n-2k+1)}{(2k)!!}$$ permutations which is roughly the above order. I also guess this case should dominate other cases when $k+1\leq h<2k$, which results in my conjecture.