Eccoci.

A unitary divisor of a positive integer $n$ is a divisor $d$ of $n$ such that $\gcd\left(d,\frac{n}{d}\right)=1$.

A bi-unitary divisor of $n$ is a divisor $d$ for which $1$ is the only unitary divisor of $d$ that is also a unitary divisor of $\frac{n}{d}$.

For example, $2$ is a bi-unitary divisor of $8$, because the unitary divisors of $2$ are $\{1,2\}$, and the unitary divisors of $8/2$ are $\{1,4\}$, with $1$ being the only unitary divisor in common.

The bi-unitary divisors of $240$ are $\{1,2,3,5,6,8,10,15,16,24,30,40,48,80,120,240\}$.

Let $P(n)$ be the product of all bi-unitary divisors of $n$. Define $Q_k(N)$ as the number of positive integers $1 \lt n \leq N$ such that $P(n)=n^k$. For example, $Q_2\left(10^2\right)=51$ and $Q_6\left(10^6\right)=6189$.

Find $\sum_{k=2}^{10}Q_k\left(10^{12}\right)$.