Theta bound of the following function Suppose the function $f(n)$ is defined as the number of $\ast$ printed in the following code

for(int i = 0; i&lt;n; i++) {
    for(int j = i; j&lt;n; j++) {
        for(int k = j + 200; k&lt;2n; k++) {
            for( int r = k^2; r&lt;n; r++) {
                print(*);
            }
        }
    }
}


I suspect the answer is $\Theta(n^{2.5})$, by running the code at $n=5000$ and $n=50000$, but I do not have a rigid proof. 

Can anyone enlighten me? Thank you!
First, note that  we print a * for every tuple $(i,j,k,r)$ such that
$0 \leq  i \leq j \leq k-200 \leq \sqrt{r}-200\leq \sqrt{n}-200$ 

For every $i$ and $j$ there are at most $(\sqrt{n}-200)$ options, for $k$ it is $\sqrt{n}$ and for $r$ it is $n$. Thus the upper bound for the number of * is $(\sqrt{n}-200)\cdot (\sqrt{n}-200) \cdot \sqrt{n} \cdot  n\in O(n^{2.5})$.

To compute the lower bound, we can take $i$'s in the range between $0$ and $\frac{\sqrt{0.5n}-200}{2}$, $j's$  between $\frac{\sqrt{0.5n}-200}{2}+1$ and $\sqrt{0.5n}$, $k$'s between $\sqrt{0.5n}+1$ and $\sqrt{0.75n}$, and finally, $r's$ between $0.75n+1$ to $n$. Thus the number of options for a tuple $(i,j,k,r)$ is more than $(\frac{\sqrt{0.5n}-200}{2})^2 \cdot (\sqrt{0.75n}-\sqrt{0.5n})\cdot 0.25n \in \Omega(n^{2.5})$.

Thus the answer is $\Theta(n^{2.5})$