Grover search on three qubits

The register has three qubits in $\lvert000\rangle$; qubit $0$ is the most significant bit, so $\lvert b_0b_1b_2\rangle$ has index $4b_0+2b_1+b_2$.

The circuit already contains the oracle placeholder $O_f$ twice. In the circuit editor it is the statement Of on its own line; keep it exactly twice, once per Grover iteration. The grader replaces both copies by the diagonal matrix $O_f=I_8-2\lvert w\rangle\langle w\rvert$, i.e. the identity with the single entry $(w,w)$ changed to $-1$.

Allowed gates: $H$, $X$, $Z$, $\mathrm{CZ}$, $\mathrm{CCZ}$, $\mathrm{CNOT}$, $\mathrm{CCX}$; at most $32$ operations including both Of statements; no measurements.

Build $H^{\otimes3}$ followed by two copies of $[\,O_f;\ \text{diffuser}\,]$. The diffuser $D=H^{\otimes3}(2\lvert000\rangle\langle000\rvert-I)H^{\otimes3}$ equals, up to the global phase $-1$, the gate sequence $H^{\otimes3}X^{\otimes3}\,\mathrm{CCZ}\,X^{\otimes3}H^{\otimes3}$, because $X^{\otimes3}\,\mathrm{CCZ}\,X^{\otimes3}=I-2\lvert000\rangle\langle000\rvert$.

grover-cases (three oracles, $w\in\{101,010,111\}$): the full distribution on qubits $[0,1,2]$ must equal $P(w)=\tfrac{121}{128}=0.9453125$ and $P(y)=\tfrac1{128}=0.0078125$ for every $y\neq w$, within $10^{-6}$.