Amplitude-encode the vector (1, 2, 3, 4)

Starting from $\lvert00\rangle$, prepare the normalised amplitude encoding of $x=(x_0,x_1,x_2,x_3)=(1,2,3,4)$:

$\lvert x\rangle=\frac{1}{\lVert x\rVert}\sum_{i=0}^{3}x_i\lvert i\rangle=\frac{1}{\sqrt{30}}\big(1\cdot\lvert00\rangle+2\cdot\lvert01\rangle+3\cdot\lvert10\rangle+4\cdot\lvert11\rangle\big),\qquad\lVert x\rVert=\sqrt{1+4+9+16}=\sqrt{30}.$

The index of $\lvert q_0q_1\rangle$ is $k=2q_0+q_1$: qubit 0 is the most significant bit, so $x_0,x_1$ sit in the half $q_0=0$ and $x_2,x_3$ in the half $q_0=1$.

Allowed gates: $R_Y(\theta)=e^{-i\theta Y/2}=\begin{pmatrix}\cos\frac\theta2&-\sin\frac\theta2\\\sin\frac\theta2&\cos\frac\theta2\end{pmatrix}$ and $\mathrm{CNOT}$, at most five gates in total. Angles are entered in radians. The grader computes the fidelity $F=\lvert\langle x\vert\psi\rangle\rvert^2$ of your final state $\lvert\psi\rangle$ and requires $F\ge1-10^{-6}$, so give each angle to at least seven significant digits.