There are $n$ students in a classroom, this classroom can be seen as a Cuboid in which the students are looking in the direction of the negative $z$-axis. The positive $x$-axis is to their right and the positive $y$-axis points upward. The wall in front of the students is the plane $z = 0$, and the wall behind them is the plane $z = a$. The walls to their right and left are the planes $x = 10^9$ and $x = -10^9$, respectively.
You will be given $q$ queries of the form ($x_i, y_i, a$), meaning that there's a point on the wall behind the students at those coordinates. You need to put a rectangular mirror with minimum area on the front wall so that all $n$ students can see that point through the mirror(The mirror needs to be *axially aligned*). Find that area for each query independently.
The first line of input contains a single integer $T$ ($1 <= T <= 8000$), the number of test cases.
The first line of each test case contains two space-separated integers $n$ and $a$ ($1 <= n <= 2 times 10^5$)($2 <= a <= 10^9$), where $n$ is the number of students, and $a$ represents that the wall behind the students is located at $z = a$.
The next $n$ lines each contains two space-separated integers $x_i$ and $z_i$ ($-10^9 < x_i < 10^9$)($0 < z_i < a$), representing that the location of the $i^(t h)$ student is ($x_i, 0, z_i$). All locations are distinct.
The next line contains an integer $q$ ($1 <= q <= 2 times 10^5$), the number of queries.
The next $q$ lines each contains two space-separated integers $x_i$ and $y_i$ ($-10^9 < x_i < 10^9$)($1 <= y_i <= 10^9$), representing that the location of the point in the $i^(t h)$ query is ($x_i, y_i, a$).
The sum of $n$ over all test cases doesn't exceed $1. 5 times 10^6$. Same goes for $q$.
For each query, print a single line with the minimum area of the mirror needed. Your answer will be considered correct if its absolute or relative error does not exceed $10^(-6)$.
## Input
The first line of input contains a single integer $T$ ($1 <= T <= 8000$), the number of test cases.The first line of each test case contains two space-separated integers $n$ and $a$ ($1 <= n <= 2 times 10^5$)($2 <= a <= 10^9$), where $n$ is the number of students, and $a$ represents that the wall behind the students is located at $z = a$.The next $n$ lines each contains two space-separated integers $x_i$ and $z_i$ ($-10^9 < x_i < 10^9$)($0 < z_i < a$), representing that the location of the $i^(t h)$ student is ($x_i, 0, z_i$). All locations are distinct.The next line contains an integer $q$ ($1 <= q <= 2 times 10^5$), the number of queries.The next $q$ lines each contains two space-separated integers $x_i$ and $y_i$ ($-10^9 < x_i < 10^9$)($1 <= y_i <= 10^9$), representing that the location of the point in the $i^(t h)$ query is ($x_i, y_i, a$).The sum of $n$ over all test cases doesn't exceed $1. 5 times 10^6$. Same goes for $q$.
## Output
For each query, print a single line with the minimum area of the mirror needed. Your answer will be considered correct if its absolute or relative error does not exceed $10^(-6)$.
[samples]
**Definitions**
Let $ T \in \mathbb{Z} $ be the number of test cases.
For each test case:
- Let $ n \in \mathbb{Z} $ be the number of students.
- Let $ a \in \mathbb{R} $ be the $ z $-coordinate of the back wall ($ z = a $).
- Let $ S = \{ (x_i, z_i) \mid i \in \{1, \dots, n\} \} $ be the set of student positions, where each student $ i $ is at $ (x_i, 0, z_i) $.
- Let $ Q = \{ (x_j, y_j) \mid j \in \{1, \dots, q\} \} $ be the set of query points, each at $ (x_j, y_j, a) $ on the back wall.
**Constraints**
1. $ 1 \le T \le 8000 $
2. $ 1 \le n \le 2 \times 10^5 $, $ 2 \le a \le 10^9 $
3. $ -10^9 < x_i < 10^9 $, $ 0 < z_i < a $ for all students
4. $ 1 \le q \le 2 \times 10^5 $
5. $ -10^9 < x_j < 10^9 $, $ 1 \le y_j \le 10^9 $ for all queries
6. $ \sum n \le 1.5 \times 10^6 $, $ \sum q \le 1.5 \times 10^6 $
**Objective**
For each query point $ P = (x_p, y_p, a) $, find the minimum area of an axis-aligned rectangular mirror on the front wall $ z = 0 $ such that every student can see $ P $ via reflection.
The mirror is defined by its projections on the $ x $- and $ y $-axes:
- Let $ x_{\min} = \min_{i} \left( x_i + \frac{z_i}{a - z_i} (x_p - x_i) \right) $
- Let $ x_{\max} = \max_{i} \left( x_i + \frac{z_i}{a - z_i} (x_p - x_i) \right) $
- Let $ y_{\min} = \min_{i} \left( \frac{z_i}{a - z_i} y_p \right) $
- Let $ y_{\max} = \max_{i} \left( \frac{z_i}{a - z_i} y_p \right) $
Then the minimum mirror area is:
$$
\text{Area} = (x_{\max} - x_{\min}) \cdot (y_{\max} - y_{\min})
$$