Calculating $\text{tr}(\Sigma^2 VV^T)$ for a diagonal matrix $\Sigma$

Question

Suppose $V \in \mathbb{R}^{n \times m}$ has orthonormal columns (here $m < n$). Let $\Sigma^2 = \text{diag}(\lambda_1^2,\dots,\lambda_n^2)$ where the diagonal entries satsify $\lambda_1^2 \geqslant \dots \geqslant \lambda_n^2 > 0$. I am interested in calculating $$\text{tr}(\Sigma^2 VV^T).$$ So far, I have that by the cyclic properties of the trace, $$\text{tr}(\Sigma^2 VV^T) = \text{tr}(\Sigma VV^T \Sigma) = \text{tr}(V^T\Sigma^2V) = \sum_{i=1}^m e_i^TV^T\Sigma^2Ve_i = \sum_{i=1}^m \|\Sigma v_i\|_2^2$$ where $e_i$ denotes the $i$-th standard basis vector and $v_i$ is the $i$-th column of $V$. I was trying to see if I can simplify this by using the orthonormality of the columns of $V$, but the weighting of $\Sigma$ is causing issues. Any tips on how to proceed?