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 = 15 * 10000 + 100;
const int lg = 18;
const int sq = 400;
const long long int mod = 998244353;
const long long int inf = 1e16;
int ti = 1, n, q, h[N], par[lg][N], stt[N], fi[N], sz[N], l[N], r[N], v[N],
    szv = 0;
long long int seg[4 * N], D[N], D2[N];
vector<int> st, hv;
vector<int> adj[N];
bool mark[N];
long long int pw(long long int x, long long int m) {
  if (m == 0) return 1;
  x %= mod;
  long long int y = (x * x) % mod;
  if (m % 2 == 0) return pw(y, m / 2);
  return (pw(y, m / 2) * x) % mod;
}
void add(int l, int r, int val, int b = 0, int e = n, int ind = 1) {
  if (e <= l || b >= r) return;
  if (l <= b && e <= r) {
    seg[ind] = (seg[ind] + val) % mod;
    return;
  }
  int mid = (b + e) / 2;
  add(l, r, val, b, mid, ind * 2);
  add(l, r, val, mid, e, ind * 2 + 1);
}
int get(int id, int b = 0, int e = n, int ind = 1) {
  if (b + 1 == e) return seg[ind];
  int mid = (b + e) / 2;
  if (id < mid) return (seg[ind] + get(id, b, mid, ind * 2)) % mod;
  return (seg[ind] + get(id, mid, e, ind * 2 + 1)) % mod;
}
void dfs(int u) {
  sz[u] = 1;
  stt[u] = ti;
  ti++;
  st.push_back(stt[u]);
  for (auto x : adj[u]) {
    if (x != par[0][u]) {
      h[x] = h[u] + 1;
      par[0][x] = u;
      for (int i = 1; i < lg; i++) par[i][x] = par[i - 1][par[i - 1][x]];
      dfs(x);
      sz[u] += sz[x];
    }
  }
  fi[u] = ti;
  ti++;
  if (int(adj[u].size()) > sq) hv.push_back(u);
  return;
}
inline int find_par(int x, int k) {
  for (int i = lg - 1; i >= 0; i--) {
    if ((k >> i) & 1) x = par[i][x];
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d  %d", &x, &y);
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(0);
  long long int inv = pw(n, mod - 2);
  for (int i = 0; i < n; i++) {
    l[i] = lower_bound(st.begin(), st.end(), stt[i]) - st.begin();
    r[i] = lower_bound(st.begin(), st.end(), fi[i]) - st.begin();
  }
  while (q--) {
    int ty, u;
    scanf("%d", &ty);
    if (ty == 1) {
      long long int d;
      scanf("%d %lld", &u, &d);
      u--;
      d %= mod;
      D[u] += d;
      D[u] %= mod;
      D2[u] += d;
      D2[u] %= mod;
      if (!mark[u]) v[szv++] = u;
      mark[u] = true;
    } else {
      scanf("%d", &u);
      u--;
      long long int ans = (get(l[u]) + D2[u]) % mod;
      for (int i = 0; i < szv; i++) {
        int x = v[i];
        if (x == u) continue;
        if (stt[x] < stt[u] && fi[x] > fi[u]) {
          int y = find_par(u, h[u] - h[x] - 1);
          ans += ((D[x] * (n - sz[y]) % mod) * inv) % mod;
        } else {
          ans += ((D[x] * sz[x] % mod) * inv) % mod;
        }
        ans %= mod;
      }
      printf("%lld \n", ans);
    }
    if (szv == sq) {
      for (int i = 0; i < szv; i++) {
        int x = v[i];
        long long int k = (((D[x] * sz[x]) % mod) * inv) % mod;
        seg[1] = (seg[1] + k) % mod;
        add(l[x], r[x], (mod - k) % mod);
        for (auto y : adj[x]) {
          if (y == par[0][x]) continue;
          long long int k = ((D[x] * (n - sz[y])) % mod * inv) % mod;
          add(l[y], r[y], k);
        }
        D[x] = 0;
        mark[x] = false;
      }
      szv = 0;
    }
  }
  return 0;
}