Given pedersen commitments of some elements, how to prove that the sum of only one subset of these elements is equal to the given element θ？

Assume that Prover have $n$ pedersen commitments ($V_{a_1},V_{a_2},\cdots,V_{a_n}$ where $V_{a_i}=G \cdot a_i + H \cdot r_{a_i}$) of $n$ elements $a_1,a_2,\cdots,a_n$. The Prover have another element $\theta$.

The verifier is provided with these $n$ pedersen commitments ($V_{a_1},V_{a_2},\cdots,V_{a_n}$) and $\theta$.

How can the Prover prove that the sum of $\textbf{only one}$ subset of these elements $a_1,a_2,\cdots,a_n$ can equal to $\theta$ and don't reveal these elements to verifier? (the size of this set is a constant witch smaller than $n$)

Let $a_i=2^{u_i}$ $i=1,\ldots,n$ with $u_i$ nonnegative integers. If the values $u_i$ are distinct, and $$ \theta=\sum_{i \in J} 2^{u_i} $$ then the subset $J\in \{1,\ldots,n\}$ is uniquely determined by $\theta$ by the uniqueness of writing $\theta$ in base $2.$

If they are not unique, then the decomposition is not necessarily unique since e.g., the repeated values $u_i,u_i$ could be replaced by $u_j=u_i+1.$

So if $u_i$ appears twice in the list then the list should not contain $u_i+1.$ This is because $2^3+2^3=2^4,$ for example. Define the condition $$ \textrm{Do}~u_i,u_i,u_{i}+1~ \textrm{appear in the list?}\quad (*) $$ If (*) does not hold the expression is unique. If it does hold then a new list with $n-1$ members should be formed with $u_i+1$ replacing the pair $u_i,u_i$ in the original list and the condition checked again recursively. Obviously if $u_i$ appeared $f_i\geq 2$ times in the original list it would appear $f_i-2$ times in the new list.

Then the condition $(*)$ needs to be applied to the new list, and this needs to be repeated again and again if $(*)$ holds until it no longer holds finally confirming uniqueness.