Build the three-qubit QFT

Build a three-qubit circuit whose $8\times8$ unitary is $F_{jk}=\tfrac1{\sqrt8}e^{2\pi ijk/8}$ for $j,k\in\{0,\dots,7\}$, using only $H$, $\mathrm{CP}(\varphi)$ and $\mathrm{SWAP}$, at most $7$ gates and no measurements. Qubit $0$ is the most significant bit: the input $\lvert j\rangle=\lvert j_1j_2j_3\rangle$ has $j_1$ on qubit $0$ and $j=4j_1+2j_2+j_3$. The grader computes the unitary $U$ of your circuit and the process fidelity $F_U=\lvert\mathrm{Tr}(F^\dagger U)\rvert/8$, which equals $1$ exactly when $U=e^{i\gamma}F$. The target is the product state $F\lvert j\rangle=\tfrac1{\sqrt8}\big(\lvert0\rangle+e^{2\pi i\,0.j_3}\lvert1\rangle\big)\otimes\big(\lvert0\rangle+e^{2\pi i\,0.j_2j_3}\lvert1\rangle\big)\otimes\big(\lvert0\rangle+e^{2\pi i\,0.j_1j_2j_3}\lvert1\rangle\big)$.