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>
const int Mod = 998244353;
int quipow(int a, int b) {
  int ret = 1;
  for (; b; b >>= 1, a = (long long)a * a % Mod)
    if (b & 1) ret = (long long)ret * a % Mod;
  return ret;
}
struct edge {
  int to, nxt;
} e[150005 << 1];
int n, q;
int h[150005], cnt;
int gson[150005], dfn[150005], hdfs[150005], siz[150005], top[150005],
    fa[150005];
int T[150005 << 2], val[150005];
void ins(int x, int y) {
  e[++cnt].nxt = h[x];
  h[x] = cnt;
  e[cnt].to = y;
}
void dfs1(int x, int f) {
  siz[x] = 1;
  fa[x] = f;
  for (int i = h[x]; i; i = e[i].nxt)
    if (e[i].to != f) {
      dfs1(e[i].to, x);
      siz[x] += siz[e[i].to];
      if (siz[gson[x]] < siz[e[i].to]) gson[x] = e[i].to;
    }
}
void dfs2(int x, int t) {
  dfn[x] = ++cnt;
  hdfs[cnt] = x;
  top[x] = t;
  if (!gson[x]) return;
  dfs2(gson[x], t);
  for (int i = h[x]; i; i = e[i].nxt)
    if (e[i].to != fa[x] && e[i].to != gson[x]) dfs2(e[i].to, e[i].to);
}
void add(int& x, int y) {
  x += y;
  x >= Mod ? x -= Mod : 0;
  x < 0 ? x += Mod : 0;
}
void modi(int x, int v) { add(T[x], v); }
void push_down(int x) {
  if (T[x]) {
    add(T[x << 1], T[x]);
    add(T[x << 1 | 1], T[x]);
    T[x] = 0;
  }
}
void modify(int x, int l, int r, int L, int R, int v) {
  if (l > R || r < L) return;
  if (l >= L && r <= R) return (void)(add(T[x], v));
  int mid = l + r >> 1;
  push_down(x);
  modify(x << 1, l, mid, L, R, v);
  modify(x << 1 | 1, mid + 1, r, L, R, v);
}
int query(int x, int l, int r, int pos) {
  if (l == r) return T[x];
  push_down(x);
  int mid = l + r >> 1;
  if (pos <= mid) return query(x << 1, l, mid, pos);
  return query(x << 1 | 1, mid + 1, r, pos);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    ins(a, b);
    ins(b, a);
  }
  cnt = 0;
  dfs1(1, 0);
  dfs2(1, 1);
  for (; q--;) {
    int opt, v, d, ans = 0;
    scanf("%d%d", &opt, &v);
    if (opt == 1) {
      scanf("%d", &d);
      modify(1, 1, n, 1, n, (long long)d * siz[v] % Mod);
      modify(1, 1, n, dfn[v], dfn[v] + siz[v] - 1,
             (long long)-d * siz[v] % Mod);
      if (gson[v])
        modify(1, 1, n, dfn[gson[v]], dfn[gson[v]] + siz[gson[v]] - 1,
               (long long)d * (n - siz[gson[v]]) % Mod);
      add(val[v], d);
    } else {
      add(ans, (long long)val[v] * n % Mod);
      add(ans, query(1, 1, n, dfn[v]));
      for (; fa[top[v]]; v = fa[top[v]])
        add(ans, (long long)(n - siz[top[v]]) * val[fa[top[v]]] % Mod);
      ans = (long long)ans * quipow(n, Mod - 2) % Mod;
      printf("%d\n", ans);
    }
  }
}