F. 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 $998\,244\,353$.

Input

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.

Output

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 $x \cdot q \equiv p \pmod{M}$.

Code

#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <stack>
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define int long long
#define endl '\n'
#define pb push_back
#define NO cout << "NO" << endl;
#define YES cout << "YES" << endl;
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
typedef vector<int> VI;
typedef vector<vector<int>> VII;
typedef pair<int,int> PII;
ll MOD = 998244353;
ll powmod(ll a,ll b) {ll res=1;a%=MOD; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%MOD;a=a*a%MOD;}return res;}
// mt19937 mrand(random_device{}()); 
const int N = 2e5 + 10;
void solve()
{
    int n, s, t; cin >> n >> s >> t;
    VII e(n + 1);
    VI sz(n + 1), de(n + 1);
    rep (i, 1, n - 1) {
    	int u, v; cin >> u >> v;
    	e[u].pb(v);
    	e[v].pb(u);
    	sz[u] ++;
    	sz[v] ++;
    }
    VII p(n + 1, VI(23));
    function<void(int, int)> dfs = [&] (int u, int Fa) {
        de[u] = de[Fa] + 1;
        p[u][0] = Fa;
        for (auto v : e[u]) {
            if (v == Fa) continue;
            dfs(v, u);
        }
    };
    dfs(t, 0);
    for (int j = 1; j < 22; j++)
        for (int i = 1; i <= n; i++) {
            p[i][j] = p[p[i][j - 1]][j-1];
        }
    function<ll(int, int)> Lca = [&] (int x, int y) -> ll {
        if (de[x] < de[y]) swap(x, y);
        for (int i = 21; i >= 0; i--) {
            if (de[p[x][i]] >= de[y]) {
                x = p[x][i];
            }
        }
        if (x == y) return x;
        for (int i = 21; i >= 0; i--) {
            if (p[x][i] != p[y][i]) {
                x = p[x][i];
                y = p[y][i];
            }
        }
        return p[x][0];
    };
    rep (i, 1, n) {
    	if (i == t) cout << 1 << ' ';
    	else
    	cout << sz[i] * (de[Lca(s, i)] - 1) % MOD << ' ';
    }
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    // int T;cin >> T;
    // while ( T -- )
    solve();
    return 0;
}