After the Wedding

Seolyeong is handing out cake to $m$ guests who attended the wedding. Every guest gets one piece, and all pieces must have the same volume.

At first the cake is a single block. Its shape is the solid formed by rotating about the $x$-axis the region enclosed by the graph of $y=f(x)$ ($x_1 \leq x \leq x_n$), the lines $x = x_1$ and $x = x_n$, and the $x$-axis. The graph of $y=f(x)$ is drawn by connecting the points $(x_1, y_1)$, $(x_2, y_2)$, $\cdots$, $(x_n, y_n)$ with line segments in this order. Here $y_i > 0 (1 \leq i \leq n)$.

To keep the pieces as neat as possible, Seolyeong always cuts the cake along a plane perpendicular to the axis of rotation. Since the cake must be divided among $m$ guests, Seolyeong makes exactly $m-1$ cuts. It can be proven that there is only one way to cut the cake under these conditions.

The force needed for a cut is proportional to the area of the cut face. If a cut produces a cross-section with area $A\pi$, the force used is $A$. Let $F_1, F_2, \cdots, F_{m-1}$ be the forces used in the $m-1$ cuts, and let $\overline{F}(m)=\frac{1}{m-1} \sum_{i=1}^{m-1} F_i$ be their average. Find

$F = \lim_{m\rightarrow\infty} \overline{F}(m)$

The first line contains an integer $n$.

The second line contains $n$ integers $x_1, x_2, \cdots, x_n$ separated by spaces.

The third line contains $n$ integers $y_1, y_2, \cdots, y_n$ separated by spaces.

These values define the graph $y=f(x)$ that gives the shape of the cake, drawn by connecting $(x_1, y_1)$, $(x_2, y_2)$, $\cdots$, $(x_n, y_n)$ with line segments in this order.

On the first line, print the value of $p \times q^{998\,244\,351} \mod 998\,244\,353$, where $F$ is written as an irreducible fraction $\displaystyle{F = \frac{p}{q}}$. $998\,244\,353$ is a prime. You can prove that $F$ is always a rational number.

• $2 \leq n \leq 300\,000$
• $0 \leq x_1 < x_2 < \cdots < x_n \leq 10^7$
• $1 \leq y_i \leq 10^7$