Cutkosky rules for a two-loop diagram

Going back to Cutkosky's original paper (http://aip.scitation.org/doi/10.1063/1.1703676), it is clear he derives his result via residue theorem, as QuantumDot pointed out in his comment. Therefore, it seems natural that the generalization of the Cutkosky's cutting rule would have to analogous to the formula for the residue of a pole of order higher than one. Explicitly, if the cut propagator is raised to the $n$-th power, we should substitute $$\frac{1}{D^n} \to (-2\pi i) \frac{(-1)^{n-1}}{(n-1)!} \delta^{(n-1)} (D).$$ In case $n=2$, we would then have $$\frac{1}{D^2} \to (2\pi i) \delta'(D).$$ While I have not yet checked this substitution rule in a real calculation, I suspect that it will hold.