3
= 3
{scope}
Throughout this question let $|A_0\rangle=\sum_{f(x)=0}|x\rangle$, $|A_1\rangle=\sum_{f(x)=1}|x\rangle$, and $|J\rangle=\sum_{x=0}^{N-1}|x\rangle=\sqrt N\,|e_0\rangle$. These subset sums are unnormalized: $\langle A_j|A_j\rangle=S_j$ and $\langle A_0|A_1\rangle=0$. The promise implies that $N>0$ is divisible by four. No power-of-two assumption is needed for the abstract $N$-dimensional calculation.