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;
inline int gi() {
  char c = getchar();
  int x = 0;
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  return x;
}
const int N = 2e5 + 5, Mod = 998244353;
vector<int> e[N];
int size[N], fa[N], son[N], top[N], dep[N], st[N], ed[N], tim, n, q, in, t[N],
    tag[N], all;
inline void add(int& x, int y) { x = (x + y >= Mod ? x + y - Mod : x + y); }
inline int po(int x, int y = Mod - 2) {
  int r = 1;
  while (y) {
    if (y & 1) r = (1ll * (r) * (x) % Mod);
    x = (1ll * (x) * (x) % Mod), y >>= 1;
  }
  return r;
}
void dfs1(int u) {
  size[u] = 1;
  dep[u] = dep[fa[u]] + 1;
  for (auto v : e[u])
    if (v != fa[u]) {
      fa[v] = u, dfs1(v);
      size[u] += size[v];
      if (size[v] > size[son[u]]) son[u] = v;
    }
}
void dfs2(int u, int tp) {
  top[u] = tp, st[u] = ++tim;
  if (son[u]) dfs2(son[u], tp);
  for (auto v : e[u])
    if (v != fa[u] && v != son[u]) dfs2(v, v);
  ed[u] = tim;
}
void upd(int i, int w) {
  for (; i <= n; i += (i & -i)) add(t[i], w);
}
int qry(int i) {
  int r = 0;
  for (; i; i -= (i & -i)) add(r, t[i]);
  return r;
}
int main() {
  n = gi(), q = gi(), in = po(n);
  for (int i = 1; i < n; ++i) {
    int u = gi(), v = gi();
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs1(1), dfs2(1, 1);
  while (q--) {
    int op = gi(), u = gi();
    if (op == 1) {
      int d = (1ll * (in) * (gi()) % Mod);
      add(all, (1ll * (size[u]) * (d) % Mod));
      add(tag[u], Mod - d);
      int tmp = (1ll * (n - size[u]) * (d) % Mod);
      upd(st[u], tmp), upd(ed[u] + 1, Mod - tmp);
      if (son[u]) {
        tmp = (1ll * (size[son[u]]) * (Mod - d) % Mod);
        upd(st[son[u]], tmp), upd(ed[son[u]] + 1, Mod - tmp);
      }
    } else {
      int ans = qry(st[u]);
      add(ans, all);
      for (; fa[top[u]]; u = fa[top[u]])
        add(ans, (1ll * (size[top[u]]) * (tag[fa[top[u]]]) % Mod));
      printf("%d\n", ans);
    }
  }
}