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 nmax = 150000 + 42, mod = 998244353;
long long my_pow(long long a, int b) {
  long long ret = 1;
  while (b) {
    if (b % 2) ret = ret * a % mod;
    b = b / 2;
    a = a * a % mod;
  }
  return ret;
}
long long inv(long long a) { return my_pow(a, mod - 2); }
int n, q;
vector<int> adj[nmax];
int in[nmax], out[nmax], depth[nmax], t, SZ[nmax];
long long fenwick[nmax];
void upd(int pos, long long val) {
  while (pos <= n) {
    fenwick[pos] += val;
    pos = pos + (pos & (-pos));
  }
}
long long query(int pos) {
  long long ret = 0;
  while (pos) {
    ret += fenwick[pos];
    pos = pos - (pos & (-pos));
  }
  return ret;
}
void update(int l, int r, long long val) {
  upd(l, val);
  upd(r + 1, -val);
}
int parent[nmax];
void dfs(int node, int par) {
  t++;
  in[node] = t;
  SZ[node] = 1;
  parent[node] = par;
  for (auto k : adj[node])
    if (k != par) {
      dfs(k, node);
      SZ[node] += SZ[k];
    }
  out[node] = t;
}
int hsh[nmax];
void dfs_renum(int node, int par) {
  t++;
  hsh[node] = t;
  vector<pair<int, int> > go = {};
  for (auto k : adj[node])
    if (k != par) go.push_back({SZ[k], k});
  sort(go.begin(), go.end());
  reverse(go.begin(), go.end());
  for (auto k : go) dfs_renum(k.second, node);
}
vector<pair<int, int> > edges = {};
void renum() {
  t = 0;
  dfs(1, 1);
  t = 0;
  dfs_renum(1, 1);
  for (int i = 1; i <= n; i++)
    for (auto j : adj[i]) edges.push_back({hsh[i], hsh[j]});
  for (int i = 1; i <= n; i++) adj[i] = {};
  for (auto k : edges) adj[k.first].push_back(k.second);
  t = 0;
}
vector<pair<pair<int, int>, int> > there[nmax];
int main() {
  scanf("%i%i", &n, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%i%i", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  renum();
  for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
  dfs(1, 1);
  for (int i = 1; i <= n; i++) {
    map<int, pair<int, int> > mem = {};
    for (auto k : adj[i])
      if (SZ[i] > SZ[k]) {
        if (mem.count(SZ[k]) == 0)
          mem[SZ[k]] = {in[k], out[k]};
        else
          mem[SZ[k]] = {min(mem[SZ[k]].first, in[k]),
                        max(mem[SZ[k]].second, out[k])};
      }
    for (auto k : mem) there[i].push_back({k.second, k.first});
  }
  for (int i = 1; i <= q; i++) {
    int type;
    scanf("%i", &type);
    if (type == 1) {
      int node, val;
      scanf("%i%i", &node, &val);
      node = hsh[node];
      update(1, n, 1LL * val * n);
      if (node != 1) {
        update(1, n, -1LL * (n - SZ[node]) * val);
        update(in[node], out[node], 1LL * (n - SZ[node]) * val);
      }
      for (auto k : there[node])
        update(k.first.first, k.first.second, -1LL * k.second * val);
    } else {
      int node;
      scanf("%i", &node);
      node = hsh[node];
      printf("%lld\n", query(in[node]) % mod * inv(n) % mod);
    }
  }
  return 0;
}