SWAP from three CNOTs

Using only CNOT gates (at most three, in either direction), build a two-qubit circuit whose unitary equals $\mathrm{SWAP}=\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix},$ that is, $\mathrm{SWAP}\,\lvert a\rangle\lvert b\rangle=\lvert b\rangle\lvert a\rangle$ for all $a,b\in\{0,1\}$. Qubit 0 is the most significant bit, so the basis order is $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$. The grader computes your circuit unitary $U$ and the process fidelity $F_U=\lvert\mathrm{Tr}(\mathrm{SWAP}^\dagger U)\rvert/4$, which equals $1$ exactly when $U=e^{i\gamma}\,\mathrm{SWAP}$ for some real $\gamma$. Measurements are not allowed.