SVD decomposition of $A(B+\lambda I)^{-1}A^T$ given SVD of $AB^{-1}A^T$

Question

Let $C=AB^{-1}A^T$, where $B$ is symmetric, positive definite and invertible. Define $C_{\lambda} = A(B+\lambda I)^{-1}A^T$, $\lambda>0$ assume we know the SVD decomposition of $C$ what can be said about the SVD of $C_{\lambda}$?

My dream would be if $\sigma_i$ is a singular value of $C$, then $\sigma_i + \lambda$ is a singular value of $C_{\lambda}$ or something like this, but I don't know how to procede. Can someone help me, please?