Problem: Hanh is a famous biologist. He loves growing trees and doing experiments on his own garden.

One day, he got a tree consisting of n vertices. Vertices are numbered from 1 to n. A tree with n vertices is an undirected connected graph with n-1 edges. Initially, Hanh sets the value of every vertex to 0.

Now, Hanh performs q operations, each is either of the following types: 

  * Type 1: Hanh selects a vertex v and an integer d. Then he chooses some vertex r uniformly at random, lists all vertices u such that the path from r to u passes through v. Hanh then increases the value of all such vertices u by d. 
  * Type 2: Hanh selects a vertex v and calculates the expected value of v. 



Since Hanh is good at biology but not math, he needs your help on these operations.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 150 000) — the number of vertices on Hanh's tree and the number of operations he performs.

Each of the next n - 1 lines contains two integers u and v (1 ≤ u, v ≤ n), denoting that there is an edge connecting two vertices u and v. It is guaranteed that these n - 1 edges form a tree.

Each of the last q lines describes an operation in either formats: 

  * 1 v d (1 ≤ v ≤ n, 0 ≤ d ≤ 10^7), representing a first-type operation. 
  * 2 v (1 ≤ v ≤ n), representing a second-type operation. 



It is guaranteed that there is at least one query of the second type.

Output

For each operation of the second type, write the expected value on a single line. 

Let M = 998244353, it can be shown that the expected value can be expressed as an irreducible fraction p/q, where p and q are integers and q not ≡ 0 \pmod{M}. Output the integer equal to p ⋅ q^{-1} mod M. In other words, output such an integer x that 0 ≤ x < M and x ⋅ q ≡ p \pmod{M}.

Example

Input


5 12
1 2
1 3
2 4
2 5
1 1 1
2 1
2 2
2 3
2 4
2 5
1 2 2
2 1
2 2
2 3
2 4
2 5


Output


1
199648871
399297742
199648871
199648871
598946614
199648873
2
2
2

Note

The image below shows the tree in the example:

<image>

For the first query, where v = 1 and d = 1: 

  * If r = 1, the values of all vertices get increased. 
  * If r = 2, the values of vertices 1 and 3 get increased. 
  * If r = 3, the values of vertices 1, 2, 4 and 5 get increased. 
  * If r = 4, the values of vertices 1 and 3 get increased. 
  * If r = 5, the values of vertices 1 and 3 get increased. 



Hence, the expected values of all vertices after this query are (1, 0.4, 0.8, 0.4, 0.4).

For the second query, where v = 2 and d = 2: 

  * If r = 1, the values of vertices 2, 4 and 5 get increased. 
  * If r = 2, the values of all vertices get increased. 
  * If r = 3, the values of vertices 2, 4 and 5 get increased. 
  * If r = 4, the values of vertices 1, 2, 3 and 5 get increased. 
  * If r = 5, the values of vertices 1, 2, 3 and 4 get increased. 



Hence, the expected values of all vertices after this query are (2.2, 2.4, 2, 2, 2).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const int N = 160000;
const int M = 610000;
const int L = 300;
int n, q;
vector<int> g[N];
int sz[N];
long long inv;
long long pow_mod(long long a, long long e) {
  long long res = 1;
  for (; e; a = a * a % mod, e >>= 1)
    if (e & 1) res = res * a % mod;
  return res;
}
long long coef[N];
int T;
int l[N], r[N], son[N];
long long sum1[M], dis[N];
int ff[N];
int father[N];
void dfs(int x, int fa) {
  sz[x] = 1;
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i] != fa) {
      dfs(g[x][i], x), sz[x] += sz[g[x][i]];
      if (sz[g[x][i]] > sz[son[x]]) son[x] = g[x][i];
    }
  coef[x] = sz[x] * inv % mod;
}
void dfs2(int x, int fa, int ant) {
  l[x] = ++T;
  ff[x] = fa;
  father[x] = ant;
  if (son[x]) dfs2(son[x], x, ant);
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i] != fa && g[x][i] != son[x]) dfs2(g[x][i], x, g[x][i]);
  r[x] = T;
}
void change(int k, int l, int r, int a, int b, long long c) {
  if (l == a && r == b) {
    sum1[k] = (sum1[k] + c) % mod;
    return;
  }
  int mid = (l + r) / 2;
  if (b <= mid)
    change(k << 1, l, mid, a, b, c);
  else if (a > mid)
    change(k << 1 | 1, mid + 1, r, a, b, c);
  else
    change(k << 1, l, mid, a, mid, c),
        change(k << 1 | 1, mid + 1, r, mid + 1, b, c);
}
long long ask(int k, int l, int r, int a) {
  if (l == r) return sum1[k];
  int mid = (l + r) / 2;
  if (a <= mid)
    return (sum1[k] + ask(k << 1, l, mid, a)) % mod;
  else
    return (sum1[k] + ask(k << 1 | 1, mid + 1, r, a)) % mod;
}
int main() {
  scanf("%d %d", &n, &q);
  inv = pow_mod(n, mod - 2);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  dfs2(1, 0, 1);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int v;
      long long d;
      scanf("%d %I64d", &v, &d);
      change(1, 1, n, 1, n, d * coef[v] % mod);
      change(1, 1, n, l[v], r[v], d),
          change(1, 1, n, l[v], r[v], mod - (d * coef[v] % mod));
      if (son[v])
        change(1, 1, n, l[son[v]], r[son[v]], mod - (d * coef[son[v]] % mod));
      dis[v] = (dis[v] + d) % mod;
    } else {
      int x;
      scanf("%d", &x);
      long long ans = ask(1, 1, n, l[x]);
      int now = father[x];
      while (now) {
        long long tmp = dis[ff[now]] * coef[now] % mod;
        ans = (ans - tmp + mod) % mod;
        now = father[ff[now]];
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}