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\textit{Problem author: Alexey Mikhnenko}

\textit{Problem developer: Viktor Romanenko}
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\textbf{Formal Problem Statement}

Given $n$ segments on a line. The following operation is allowed: select two segments $(l_i, r_i)$ and $(l_j, r_j)$ and arbitrarily form two new segments from their four endpoints.

There are $q$ queries $k_i$. For each query, it is required to determine the minimum number of operations needed to ensure that there exists a subset of $k_i$ pairwise intersecting segments.

\textbf{Main Idea}

Let all segments of the desired set intersect at some point $x$. Assume that:
\begin{itemize}
\item $c$ is the number of segments already covering point $x$,
\item $l$ is the number of segments located strictly to the left of $x$,
\item $r$ is the number of segments located strictly to the right of $x$.
\end{itemize}

Note that with one operation, one can take one left segment and one right segment and rearrange them so that both intersect point $x$. Therefore, with one operation, the number of segments intersecting $x$ can be increased by two.

Thus, the maximum number of segments that can be made to intersect at point $x$ is equal to
\[
c + 2 \cdot \min(l, r).
\]

To achieve this value, $\min(l, r)$ operations will be required. If an intermediate number $k_i$ is needed, it is sufficient to perform
\[
\left\lceil \frac{k_i - c}{2} \right\rceil
\]
operations.

\textbf{Subgroup 3. Segments Do Not Intersect}

In this case, for any point, $c \le 1$. Therefore, it is convenient to consider a point inside the central segment.

For any point inside such a segment, $c = 1$, hence the answer for any $k_i$ is calculated using the formula
\[
\left\lceil \frac{k_i - 1}{2} \right\rceil.
\]

If the number of segments is even and $k_i = n$, then the intersection point must be located between the two central segments, where $c = 0$. In this case, the answer is equal to $n/2$. Since $n$ is even, this value is equivalent to the formula above.

\textbf{Subgroup 4. $k_i = n$}

If it is required to make all segments intersect, the answer point must be located at a position where the number of segments is strictly equal on both the left and right.

Such a point can be found using a sweep line algorithm. After finding the corresponding values of $c$, $l$, and $r$, the answer is calculated using the previously mentioned formula.

\textbf{Solution for Subgroups 1, 2, 4, 5}

We will perform a sweep line. For each opening boundary of a segment and each closing boundary, we will fix the values:
\begin{itemize}
\item $c$ --- the number of segments covering the current point,
\item $\min(l, r)$ --- the minimum number of segments strictly to the left and strictly to the right.
\end{itemize}

When processing a closing boundary, the corresponding segment is no longer counted in $c$, but is considered to be strictly to the left, since the intersection point may be outside its boundaries.

For each such position, the minimum number of operations can be calculated using the formula
\[
\left\lceil \frac{k_i - c}{2} \right\rceil,
\]
provided that
\[
k_i \le c + 2 \cdot \min(l, r).
\]

The minimum across all such positions gives the answer for subgroups 1, 2, 4, and 5.

\textbf{Complete Solution}

It remains to optimize the computation of answers.

For each $k$ from $1$ to $n$, we will precompute the maximum value of $c$ such that it is possible to obtain $k$ intersecting segments. We denote this quantity as $best[k]$.

During the sweep line, for each position, we compute the value
\[
c + 2 \cdot \min(l, r),
\]
which shows the maximum number of segments intersecting at this point after operations. Then we perform the update
\[
best[c + 2 \cdot \min(l, r)] = c.
\]

After completing the sweep line, we will traverse the $best$ array from right to left and propagate the maximum:
\[
best[i] = \max(best[i], best[i + 1]).
\]

Now $best[k]$ contains the maximum number of segments already covering the point from which $k$ intersecting segments can be obtained.

The answer to the query $k_i$ is computed using the formula
\[
\left\lceil \frac{k_i - best[k_i]}{2} \right\rceil.
\]

In total, the complexity is $O(n \log n)$ for the preprocessing due to sorting and $O(1)$ for each query