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 M = 998244353;
const int N = 2e5 + 7, RT = 400;
int power(int a, int p) {
  if (p == 0) return 1;
  int ans = power(a, p / 2);
  ans = (1LL * ans * ans) % M;
  if (p % 2) ans = (1LL * a * ans) % M;
  return ans;
}
int inv[N];
void pre() {
  for (int i = 1; i < N; i++) {
    inv[i] = power(i, M - 2);
    assert((1LL * i * inv[i]) % M == 1);
  }
}
vector<int> adj[N];
int lev[N], st[N], en[N], sz[N], deg[N], par[N];
int tim = 0;
void dfs(int u, int p) {
  par[u] = p;
  st[u] = ++tim;
  lev[u] = lev[p] + 1;
  sz[u] = 1;
  for (int v : adj[u])
    if (v != p) {
      dfs(v, u);
      sz[u] += sz[v];
    }
  en[u] = tim;
}
int sum[N / RT + 10], a[N];
void add(int i, int x) {
  sum[i / RT] += x;
  if (sum[i / RT] >= M) sum[i / RT] -= M;
  a[i] += x;
  if (a[i] >= M) a[i] -= M;
}
int query(int r) {
  int br = r / RT;
  long long ans = 0;
  for (int i = br * RT; i <= r; i++) ans += a[i];
  for (int i = 0; i < br; i++) ans += sum[i];
  return ans % M;
}
bool heavy[N];
int lz[N];
vector<int> heavynodes;
int n;
void updbrute(int u, int x) {
  for (int v : adj[u]) {
    if (v == par[u]) {
      int rem = sz[u];
      int y = (1LL * x * rem) % M;
      add(1, y);
      add(st[u], M - y);
      add(en[u] + 1, y);
      add(n + 1, M - y);
      continue;
    }
    int rem = n - sz[v];
    int y = (1LL * x * rem) % M;
    add(st[v], y);
    add(en[v] + 1, M - y);
  }
  int y = (1LL * x * n) % M;
  add(st[u], y);
  add(st[u] + 1, M - y);
}
vector<int> sizes[N];
int dfs1(int u, int p) {
  int ans = 1;
  for (int v : adj[u])
    if (v != p) ans += dfs1(v, u);
  return ans;
}
void dfs2(int u, int p, int x, int id) {
  sizes[id][u] = x;
  for (int v : adj[u])
    if (v != p) dfs2(v, u, x, id);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  pre();
  int q;
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    if (deg[i] > 500) {
      heavy[i] = true;
      heavynodes.push_back(i);
      sizes[i].resize(n + 1);
      sizes[i][i] = n;
      for (int v : adj[i]) {
        int x = dfs1(v, i);
        dfs2(v, i, n - x, i);
      }
    }
  }
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int u, d;
      cin >> u >> d;
      if (!heavy[u]) updbrute(u, d);
      lz[u] += d;
      if (lz[u] >= M) lz[u] -= M;
    } else {
      int u;
      cin >> u;
      long long ans = query(st[u]);
      for (int v : heavynodes) ans = (ans + 1LL * lz[v] * sizes[v][u]) % M;
      cout << (ans * inv[n]) % M << "\n";
    }
  }
}