Send the message 11 with one qubit

The first two operations, $H$ on qubit $0$ and $\mathrm{CNOT}$ with control $0$ and target $1$, are locked: they create the shared pair $\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)=\tfrac1{\sqrt2}(1,0,0,1)^T,$ with qubit $0$ (Alice, the most significant bit) and qubit $1$ (Bob). Append at most four gates from $\{X,Z,\mathrm{CNOT},H\}$, no measurements:

1. Encode the message $b_1b_2=11$ by applying $Z^{b_1}X^{b_2}$ to Alice's qubit $0$, i.e. $X$ and then $Z$ on qubit $0$. Alice may only act on her own qubit; qubit $1$ accepts nothing except the decoding $\mathrm{CNOT}$.
2. Decode with $(H\otimes I)\,\mathrm{CNOT}_{0\to1}$: first $\mathrm{CNOT}$ with control $0$ and target $1$, then $H$ on qubit $0$. This is $B^\dagger$ for the Bell circuit $B=\mathrm{CNOT}(H\otimes I)$.

At most one two-qubit gate may follow the locked prefix. Three checks are run, all with tolerance $10^{-6}$: decoded-11 compares $(p_{00},p_{01},p_{10},p_{11})$ with $(0,0,0,1)$; first-bit compares the marginal of qubit $0$ with $(p_0,p_1)=(0,1)$; second-bit compares the marginal of qubit $1$ with $(0,1)$.