I need to prove that given vector of commitments of length N contains N-1 commitments to zero (and one to an arbitrary number). More formally, given vector: $$\textbf{a} = \begin{bmatrix} C(0, r_1) & C(0, r_2) & C(x, r_3) & ... & C(0, r_N) \end{bmatrix}$$ I want to prove that there is exactly one such commitment, that commits to x, rather than 0.
I've seen the opposite to what I need to prove in One-out-of-Many Proofs paper, but still can't come up with what I need.

I need to prove that given vector of commitments of length N contains N-1 commitments to zero (and one to an arbitrary number). More formally, given vector: $$\textbf{a} = \begin{bmatrix} C(0, r_1) & C(0, r_2) & C(x, r_3) & ... & C(0, r_N) \end{bmatrix}$$ I want to prove that there is exactly one such commitment, that commits to x, rather than 0. Note that x is also private.
I've seen the opposite to what I need to prove in One-out-of-Many Proofs paper, but still can't come up with what I need.