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 = 1000100;
const int M = 2 * N;
const int md = 998244353;
const int BLOCK = 380;
const int PW = 20;
const int oo = 2e9;
const long long OO = 1e18;
vector<int> g[N];
int fen[N], id[N], siz[N], top[N], n, q, vl[N], pr[N], son[N], tt = 0, dv,
                                                               gl = 0;
int sum(int x, int y) { return (x + y) % md; }
int mult(int x, int y) { return (1ll * x * y) % md; }
int binpow(int x, int pw) {
  if (pw == 1) return x;
  if (pw == 0) return 1;
  if (pw & 1) {
    return mult(binpow(x, pw - 1), x);
  } else {
    int res = binpow(x, pw >> 1);
    return mult(res, res);
  }
}
void dfs(int v, int p) {
  siz[v] = 1;
  pr[v] = p;
  for (int u : g[v]) {
    if (u == p) continue;
    dfs(u, v);
    siz[v] += siz[u];
    if (!son[v] || siz[son[v]] < siz[u]) son[v] = u;
  }
}
void rec(int v, int tp) {
  top[v] = tp;
  id[v] = tt++;
  if (son[v]) rec(son[v], tp);
  for (int u : g[v]) {
    if (u == pr[v]) continue;
    if (u == son[v]) continue;
    rec(u, u);
  }
}
void update(int v, int x) {
  for (; v < n; v = (v | (v + 1))) fen[v] = sum(fen[v], x);
}
int get(int v) {
  int res = 0;
  for (; v >= 0; v = (v & (v + 1)) - 1) res = sum(res, fen[v]);
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dv = binpow(n, md - 2);
  dfs(0, -1);
  rec(0, 0);
  for (; q; q--) {
    int tp, v;
    cin >> tp >> v;
    v--;
    if (tp == 1) {
      int d;
      cin >> d;
      vl[v] = sum(vl[v], d);
      int cur = mult(d, siz[v]);
      gl = sum(gl, cur);
      update(id[v], md - cur);
      update(id[v] + siz[v], +cur);
      if (!son[v]) continue;
      cur = mult(d, n - siz[son[v]]);
      update(id[son[v]], cur);
      update(id[son[v]] + siz[son[v]], md - cur);
    } else {
      int res = get(id[v]);
      res = sum(res, gl);
      res = sum(res, mult(vl[v], n));
      for (int u = top[v]; pr[u] >= 0; u = top[pr[u]])
        res = sum(res, mult(n - siz[u], vl[pr[u]]));
      cout << mult(dv, res) << '\n';
    }
  }
}