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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int maxn = 150100 + 7;
const int mod = 998244353;
const int INF = 1e9 + 7;
const int mlog = 20;
const int SQ = 95;
int n, q;
inline long long pw(long long a, long long b) {
  long long res = 1ll;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = a * a % mod;
    b /= 2;
  }
  return res;
}
vector<int> adj[maxn];
vector<pair<int, int> > g[maxn];
int sz[maxn];
int timer = 0;
int st[maxn], en[maxn];
long long fen[maxn];
long long frac[maxn];
bool cmp(int u, int v) { return sz[u] < sz[v]; }
inline void UPDATE(int ind, long long x) {
  for (ind++; ind < maxn; ind += ind & -ind) fen[ind] = (fen[ind] + x);
}
inline void ADD(int l, int r, long long x) { UPDATE(l, x), UPDATE(r + 1, -x); }
inline long long GET(int ind) {
  long long res = 0;
  for (ind++; ind > 0; ind -= ind & -ind) {
    res = (res + fen[ind]) % mod;
  }
  return res;
}
void dfs(int v, int par = -1) {
  st[v] = timer++;
  sz[v] = 1;
  for (int u : adj[v]) {
    if (u == par) continue;
    dfs(u, v);
    sz[v] += sz[u];
  }
  en[v] = timer;
}
int32_t main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) frac[i] = ((i)*pw((n), mod - 2) % mod);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v), adj[v].push_back(u);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end(), cmp);
  timer = 0;
  dfs(1);
  for (int v = 1; v <= n; v++) {
    int ind = 0;
    while (ind < adj[v].size() - (v != 1)) {
      int pos = upper_bound(adj[v].begin() + ind + 1, adj[v].end(), adj[v][ind],
                            cmp) -
                adj[v].begin() - 1;
      g[v].push_back({ind, pos});
      ind = pos + 1;
    }
  }
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int v, d;
      scanf("%d%d", &v, &d);
      ADD(st[v], st[v], d);
      long long val = d * frac[sz[v]] % mod;
      ADD(0, st[v] - 1, val);
      ADD(en[v], n - 1, val);
      for (auto p : g[v]) {
        int ind = p.first;
        int pos = p.second;
        long long val = d * frac[n - sz[adj[v][ind]]] % mod;
        ADD(st[adj[v][ind]], en[adj[v][pos]] - 1, val);
      }
    }
    if (type == 2) {
      int v;
      scanf("%d", &v);
      long long ans = GET(st[v]);
      printf("%lld\n", ans);
    }
  }
  return 0;
}