CF1823F.Random Walk

You are given a tree consisting of $n$ vertices and $n - 1$ edges, and each vertex $v$ has a counter $c(v)$ assigned to it.

Initially, there is a chip placed at vertex $s$ and all counters, except $c(s)$ , are set to $0$ ; $c(s)$ is set to $1$ .

Your goal is to place the chip at vertex $t$ . You can achieve it by a series of moves. Suppose right now the chip is placed at the vertex $v$ . In one move, you do the following:

1. choose one of neighbors $to$ of vertex $v$ uniformly at random ( $to$ is neighbor of $v$ if and only if there is an edge $\{v, to\}$ in the tree);
2. move the chip to vertex $to$ and increase $c(to)$ by $1$ ;

You'll repeat the move above until you reach the vertex $t$ .

For each vertex $v$ calculate the expected value of $c(v)$ modulo $998\,244\,353$ .

The first line contains three integers $n$ , $s$ and $t$ ( $2 \le n \le 2 \cdot 10^5$ ; $1 \le s, t \le n$ ; $s \neq t$ ) — number of vertices in the tree and the starting and finishing vertices.

Next $n - 1$ lines contain edges of the tree: one edge per line. The $i$ -th line contains two integers $u_i$ and $v_i$ ( $1 \le u_i, v_i \le n$ ; $u_i \neq v_i$ ), denoting the edge between the nodes $u_i$ and $v_i$ .

It's guaranteed that the given edges form a tree.

Print $n$ numbers: expected values of $c(v)$ modulo $998\,244\,353$ for each $v$ from $1$ to $n$ .

Formally, let $M = 998\,244\,353$ . It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$ , where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$ . Output the integer equal to $p \cdot q^{-1} \bmod M$ . In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$ .

The tree from the first example is shown below:

Let's calculate expected value $E[c(1)]$ : - $P(c(1) = 0) = 0$ , since $c(1)$ is set to $1$ from the start.

• $P(c(1) = 1) = \frac{1}{2}$ , since there is the only one series of moves that leads $c(1) = 1$ . It's $1 \rightarrow 2 \rightarrow 3$ with probability $1 \cdot \frac{1}{2}$ .
• $P(c(1) = 2) = \frac{1}{4}$ : the only path is $1 \rightarrow_{1} 2 \rightarrow_{0.5} 1 \rightarrow_{1} 2 \rightarrow_{0.5} 3$ .
• $P(c(1) = 3) = \frac{1}{8}$ : the only path is $1 \rightarrow_{1} 2 \rightarrow_{0.5} 1 \rightarrow_{1} 2 \rightarrow_{0.5} 1 \rightarrow_{1} 2 \rightarrow_{0.5} 3$ .
• $P(c(1) = i) = \frac{1}{2^i}$ in general case.

As a result, $E[c(1)] = \sum\limits_{i=1}^{\infty}{i \frac{1}{2^i}} = 2$ . Image of tree in second test Image of tree in third test