\left(m l_{\rm max}\right) \right\rceil$ qubits, so the total number which must be transmitted for disentanglement is still $O (\sqrt{m})$.

In the second subcase, the number $m$ is chosen such that the $m$-th signal is most likely already in the receiver (Bob)’s hands. \[e.g. $m
< K - O (\sqrt{K})$ in the asymptotic limit.\] The receiver Bob now attempts to disentangle the remaining signals from the first $m$ quantum signals by applying a quantum shifting operation. Of course, he needs to shift some of his qubits back to Alice. This asymptotically amounts to $O (\sqrt{m})$ qubits of [*backward*]{} communication. This is a penalty that one must pay for this method. After this is done, Alice must again send her length register to Bob (after subtracting the lengths of the signals returned to her). This requires an additional $O(\log m)$ qubits.

If $m$ is chosen between $K - O(\sqrt{K})$ and $K + O(\sqrt{K})$, neither sending signals forward or backward will suffice to properly disentangle the varying lengths of the signals. One possible solution is to choose $m' > K + O(\sqrt{K})$ and perform the above procedure, sending $m'$ total signals to Bob. Then Bob decodes and returns the $m' - m$ extra 