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 sz = 2e5, mod = 998244353;
int po(int x, int y) {
  int re = 1;
  while (y) {
    if (y & 1) re = (1ll * re * x) % mod;
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return re;
}
vector<int> sv[sz];
int pr[sz], si[sz], tin[sz], timer = 0, hld[sz], su[sz];
bool comp(int a, int b) { return si[a] < si[b]; }
void dfs(int v) {
  for (int a = 0; a < sv[v].size(); a++) {
    if (sv[v][a] == pr[v]) sv[v].erase(sv[v].begin() + a);
  }
  tin[v] = timer++;
  si[v] = 1;
  for (int a = 0; a < sv[v].size(); a++) {
    int ne = sv[v][a];
    pr[ne] = v, dfs(ne), si[v] += si[ne];
  }
  sort(sv[v].begin(), sv[v].end(), comp);
}
void dfs2(int v) {
  for (int a = 0; a + 1 < sv[v].size(); a++) hld[sv[v][a]] = sv[v][a];
  if (sv[v].size()) hld[sv[v].back()] = hld[v];
  for (int a = 0; a < sv[v].size(); a++) dfs2(sv[v][a]);
}
int dot[sz * 4], add[sz * 4];
void push(int nu) {
  add[nu * 2] = (add[nu * 2] + add[nu]) % mod;
  add[nu * 2 + 1] = (add[nu * 2 + 1] + add[nu]) % mod;
  add[nu] = 0;
}
void recalc(int nu, int q) {
  dot[nu] = (dot[nu * 2] + 1ll * (q - q / 2) * add[nu * 2] + dot[nu * 2 + 1] +
             1ll * (q / 2) * add[nu * 2 + 1]) %
            mod;
}
int lg, rg, val;
void cha(int nu, int l, int r) {
  if (l >= lg and r <= rg)
    add[nu] = (add[nu] + val) % mod;
  else if (!(l > rg or r < lg)) {
    int mid = (l + r) >> 1;
    push(nu);
    cha(nu * 2, l, mid);
    cha(nu * 2 + 1, mid + 1, r);
    recalc(nu, r - l + 1);
  }
}
int p;
int que(int nu, int l, int r) {
  if (l == r)
    return (dot[nu] + add[nu]) % mod;
  else {
    int mid = (l + r) >> 1, re;
    push(nu);
    if (p <= mid)
      re = que(nu * 2, l, mid);
    else
      re = que(nu * 2 + 1, mid + 1, r);
    recalc(nu, r - l + 1);
    return re;
  }
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int a = 0; a < n - 1; a++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    sv[u].push_back(v);
    sv[v].push_back(u);
  }
  dfs(0), dfs2(0);
  int rev = po(n, mod - 2);
  for (int a = 0; a < q; a++) {
    int ty, v;
    scanf("%d%d", &ty, &v);
    v--;
    if (ty == 1) {
      int d;
      scanf("%d", &d);
      lg = 0, rg = n - 1, val = (1ll * d * ((1ll * rev * si[v]) % mod)) % mod;
      cha(1, 0, n - 1);
      lg = tin[v], rg = tin[v] + si[v] - 1, val = (mod - val) % mod;
      cha(1, 0, n - 1);
      if (sv[v].size()) {
        int u = sv[v].back();
        lg = tin[u], rg = tin[u] + si[u] - 1,
        val = (1ll * d * ((1ll * rev * (n - si[u])) % mod)) % mod;
        cha(1, 0, n - 1);
      }
      su[v] = (su[v] + d) % mod;
    } else {
      p = tin[v];
      int an = (que(1, 0, n - 1) + su[v]) % mod;
      while (v) {
        v = hld[v];
        an = (an + 1ll * su[pr[v]] * ((1ll * rev * (n - si[v])) % mod)) % mod;
        v = pr[v];
      }
      printf("%d\n", an);
    }
  }
}