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;
const int MN = 150005;
inline int gInv(int b) {
  int a = 1, e = Mod - 2;
  for (; e; e >>= 1, b = (long long)b * b % Mod)
    if (e & 1) a = (long long)a * b % Mod;
  return a;
}
int N, Q;
std::vector<int> G[MN];
int faz[MN], dep[MN], siz[MN], son[MN], top[MN], ldf[MN], rdf[MN], dfc;
void DFS0(int u, int fz) {
  dep[u] = dep[faz[u] = fz] + 1, siz[u] = 1;
  for (auto v : G[u])
    if (v != fz) {
      DFS0(v, u);
      siz[u] += siz[v];
      if (siz[son[u]] < siz[v]) son[u] = v;
    }
}
void DFS1(int u, int tp) {
  top[u] = tp, ldf[u] = ++dfc;
  if (son[u]) DFS1(son[u], tp);
  for (auto v : G[u])
    if (v != faz[u] && v != son[u]) DFS1(v, v);
  rdf[u] = dfc;
}
int B[MN];
inline void Add(int i, int x) {
  for (; i <= N; i += i & -i) B[i] -= (B[i] += x) >= Mod ? Mod : 0;
}
inline int Qur(int i) {
  int A = 0;
  for (; i; i -= i & -i) A -= (A += B[i]) >= Mod ? Mod : 0;
  return A;
}
int tag[MN], InvN, totSum;
int main() {
  scanf("%d%d", &N, &Q), InvN = gInv(N);
  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);
  }
  DFS0(1, 0), DFS1(1, 0);
  while (Q--) {
    int op, u, x;
    scanf("%d%d", &op, &u);
    if (op == 1) {
      scanf("%d", &x);
      totSum = (totSum + (long long)x * siz[u] % Mod * InvN) % Mod;
      int val = (long long)x * (N - siz[u]) % Mod * InvN % Mod;
      Add(ldf[u], val);
      Add(rdf[u] + 1, Mod - val);
      tag[u] -= x;
      if (tag[u] < 0) tag[u] += Mod;
      if (son[u]) {
        int val2 = (long long)(Mod - x) * siz[son[u]] % Mod * InvN % Mod;
        Add(ldf[son[u]], val2);
        Add(rdf[son[u]] + 1, Mod - val2);
      }
    } else {
      int Ans = Qur(ldf[u]);
      while (faz[top[u]]) {
        u = top[u];
        Ans = (Ans + (long long)tag[faz[u]] * siz[u] % Mod * InvN) % Mod;
        u = faz[u];
      }
      printf("%d\n", (totSum + Ans) % Mod);
    }
  }
  return 0;
}