Binary Wine

Think of the operation like this: for one query, we choose final upper bounds $A_i \ge a_i$ and pay $\sum (A_i-a_i)$. Then we only need to know whether some $b_i$ with $0 \le b_i \le A_i$ can XOR to $c$.

The sneaky part is that feasibility is not about XORing the $a_i$ themselves. It is about the ranges $[0,A_i]$. Big difference. Parity-only thinking gets cooked here.

The feasibility test

For every bit $k$, define

$s_k =$ the number of values $A_i$ whose bit $k$ is set.

Compare the sequence $(s_{29},s_{28},\dots,s_0)$ with the binary digits of $c$, also from high to low. The target $c$ is achievable iff this count sequence is lexicographically at least the bit sequence of $c$.

Since each bit of $c$ is only $0$ or $1$, the only real failure is: at the first differing bit, $c$ has $1$ but the count is $0$.

Why does this work? Look at the highest bit $k$ still being considered.

If more numbers have bit $k$ than $c_k$, then we have slack and all lower bits become free. If $c_k=0$, one number with bit $k$ set can simply choose any lower value below $2^k$, so the high bit stays $0$ and the lower XOR can be anything. If $c_k=1$ and at least two numbers have bit $k$, choose one as exactly $2^k$ and another as any lower value. Again, lower bits are whatever we want.

If the count equals $c_k$, the bit is forced. If both are $0$, no one can use this bit, so we just continue below it. If both are $1$, exactly one number has this bit, and it must provide the high bit of the XOR. Its remaining lower freedom is just its lower part, so we clear bit $k$ from that upper bound and continue.

If the count is less than $c_k$, then $c_k=1$ and nobody can provide that bit. Impossible.

That gives a clean high-to-low check.

Now minimize coins

We simulate that same check, but when it would fail, we pay the cheapest amount to fix it.

Maintain current residual upper bounds after all forced higher bits have been stripped. Process bits from $29$ down to $0$.

Let have be the number of residuals with bit $k$ set, and need be bit $k$ of $c$.

If have > need, stop. The current paid amount already works.

If have == need, this bit is forced, so clear bit $k$ from every residual and continue.

If have < need, then need=1 and have=0. Every residual is below $2^k$, so we must raise one of them until bit $k$ becomes $1$. Raising residual $x$ to exactly $2^k$ costs $2^k-x$, so we choose the largest $x$. After that, this bit is forced and the chosen number has lower residual $0$.

Raising above $2^k$ is not magic. Any extra amount only buys lower residual bits, and the later steps can pay for those exactly if they are actually useful. So setting it to exactly $2^k$ is enough.

Why top 40 values are enough

Doing this over all $n$ values for every query would be way too slow.

Sort $a$ descending and keep only the largest 40 values. This is safe because there are only 30 bits.

Suppose the scan reaches bit $k$. Then for every higher bit, we did not stop, meaning the count at that bit was equal to the corresponding bit of $c$. So each higher bit can account for at most one relevant large number. Across all higher bits, that is at most 30 annoying blockers.

Therefore, if bit $k$ has two available numbers, at least two of them appear among roughly the first 32 original values. If bit $k$ has none and we need the largest residual below $2^k$, that largest residual is also behind at most those 30 blockers. Keeping 40 is just a comfortable no-drama margin.

So each query costs only $40 \cdot 30$ tiny operations.

Complexity

Sorting costs $O(n \log n)$ per test case. Each query costs $O(40 \cdot 30)$, which is effectively constant. With the given limits, this is very comfortably AC.