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 int N = 1.5E5, P = 998244353;
int n, q;
vector<int> e[N];
int sum[N], sz[N], top[N], in[N], dep[N], parent[N], out[N];
int dfs_now;
int fen[N + 1];
void add(int x, int y) {
  while (x <= N) {
    fen[x] = (fen[x] + y) % P;
    x += x & -x;
  }
}
void add(int l, int r, int x) {
  add(l + 1, x);
  add(r + 1, P - x);
}
int get(int x) {
  ++x;
  int result = 0;
  while (x > 0) {
    result = (result + fen[x]) % P;
    x -= x & -x;
  }
  return result;
}
int power(int base, int exp) {
  int result = 1;
  while (exp) {
    if (exp & 1) result = 1LL * result * base % P;
    base = 1LL * base * base % P;
    exp >>= 1;
  }
  return result;
}
void dfsSz(int u) {
  sz[u] = 1;
  if (parent[u] != -1) e[u].erase(find(e[u].begin(), e[u].end(), parent[u]));
  for (int& v : e[u]) {
    dep[v] = dep[u] + 1;
    parent[v] = u;
    dfsSz(v);
    sz[u] += sz[v];
    if (sz[v] > sz[e[u][0]]) swap(v, e[u][0]);
  }
}
void dfsHLD(int u) {
  in[u] = dfs_now++;
  for (int v : e[u]) {
    top[v] = v == e[u][0] ? top[u] : v;
    dfsHLD(v);
  }
  out[u] = dfs_now;
}
void modify(int v, int d) {
  add(0, in[v], 1LL * d * sz[v] % P);
  add(out[v], n, 1LL * d * sz[v] % P);
  add(in[v], in[v] + 1, 1LL * d * n % P);
  if (!e[v].empty()) {
    int u = e[v][0];
    add(in[u], out[u], 1LL * d * (n - sz[u]) % P);
  }
  sum[v] = (sum[v] + d) % P;
}
int query(int v) {
  int result = get(in[v]);
  while (v != -1) {
    v = top[v];
    if (parent[v] != -1)
      result = (result + 1LL * sum[parent[v]] * (n - sz[v])) % P;
    v = parent[v];
  }
  return result;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> q;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    e[--u].push_back(--v);
    e[v].push_back(u);
  }
  parent[0] = -1;
  dfsSz(0);
  dfsHLD(0);
  int inv = power(n, P - 2);
  for (int i = 0; i < q; ++i) {
    int op;
    cin >> op;
    if (op == 1) {
      int v, d;
      cin >> v >> d;
      modify(--v, d);
    } else {
      int v;
      cin >> v;
      cout << 1LL * query(--v) * inv % P << "\n";
    }
  }
  return 0;
}