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;
int read() {
  int x = 0;
  bool flg = false;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') flg = true;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
  return flg ? -x : x;
}
const int Mod = 998244353;
int qpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1LL * x * x % Mod)
    if (y & 1) res = 1LL * res * x % Mod;
  return res;
}
int n, m;
int w[150010];
struct Edge {
  int to, nxt;
} edge[300010];
int cnt = 1, last[150010];
inline void addedge(int x, int y) {
  edge[++cnt] = (Edge){y, last[x]}, last[x] = cnt;
  edge[++cnt] = (Edge){x, last[y]}, last[y] = cnt;
}
int fat[150010], son[150010], siz[150010], dep[150010];
int top[150010];
void dfs1(int cur, int fat, int dep) {
  ::fat[cur] = fat;
  ::dep[cur] = dep;
  siz[cur] = 1;
  for (int i = last[cur], v; i; i = edge[i].nxt)
    if (v = edge[i].to, v ^ fat) {
      dfs1(v, cur, dep + 1);
      siz[cur] += siz[v];
      if (!son[cur] || siz[son[cur]] < siz[v]) son[cur] = v;
    }
}
int id[150010];
void dfs2(int cur, int top) {
  ::top[cur] = top;
  id[cur] = ++*id;
  if (siz[cur] == 1) return;
  dfs2(son[cur], top);
  for (int i = last[cur], v; i; i = edge[i].nxt)
    if (v = edge[i].to, (v ^ fat[cur]) && (v ^ son[cur])) dfs2(v, v);
}
int tri[150010];
void add(int x, int val) {
  for (; x <= n; x += x & -x) tri[x] = (tri[x] + val) % Mod;
}
int ask(int x) {
  int res = 0;
  for (; x; x -= x & -x) res = (res + tri[x]) % Mod;
  return res;
}
int main() {
  n = read(), m = read();
  int invn = qpow(n, Mod - 2);
  for (int i = 1; i < n; i++) addedge(read(), read());
  dfs1(1, 0, 1);
  dfs2(1, 1);
  int opt, x, y, z;
  while (m--) {
    opt = read(), x = read();
    if (opt == 1) {
      y = read();
      w[x] = (w[x] + y) % Mod;
      z = 1LL * siz[x] * y % Mod;
      add(1, z);
      add(id[x], -z);
      add(id[x] + siz[x], z);
      if (!son[x]) continue;
      z = 1LL * (n - siz[son[x]]) * y % Mod;
      add(id[son[x]], z);
      add(id[son[x]] + siz[son[x]], -z);
      continue;
    }
    int res = ask(id[x]);
    res = (res + 1LL * w[x] * n) % Mod;
    for (int i = top[x]; fat[i]; i = top[fat[i]])
      res = (res + 1LL * (n - siz[i]) * w[fat[i]]) % Mod;
    res = 1LL * res * invn % Mod;
    printf("%d\n", (res + Mod) % Mod);
  }
  return 0;
}