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 = 22;
const long long int mod = 998244353;
const long long int inf = 1e16;
int sz[N], par[N], st[N], fi[N], ti = 1;
vector<int> v, adj[N];
vector<pair<pair<int, int>, int>> g[N];
long long int D[N], fen[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 dfs1(int u) {
  sz[u] = 1;
  for (auto x : adj[u]) {
    if (x != par[u]) {
      par[x] = u;
      dfs1(x);
      sz[u] += sz[x];
    }
  }
}
void dfs2(int u) {
  st[u] = ti++;
  v.push_back(st[u]);
  for (auto x : adj[u]) {
    if (x != par[u]) {
      dfs2(x);
    }
  }
  fi[u] = ti++;
}
bool cmp(int x, int y) { return sz[x] < sz[y]; }
inline void add2(int p, long long int x) {
  for (; p; p -= p & -p) fen[p] += x;
}
inline void add(int l, int r, long long int x) {
  add2(l, -x);
  add2(r, x);
}
inline long long int get(int p) {
  long long int sum = 0;
  for (p++; p < N; p += p & -p) sum += fen[p];
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, q;
  scanf("%d %d", &n, &q);
  long long int inv = pw(n, mod - 2);
  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);
  }
  dfs1(0);
  for (int i = 0; i < n; i++) sort(adj[i].begin(), adj[i].end(), cmp);
  dfs2(0);
  for (int u = 0; u < n; u++) {
    st[u] = lower_bound(v.begin(), v.end(), st[u]) - v.begin();
    fi[u] = lower_bound(v.begin(), v.end(), fi[u]) - v.begin();
  }
  for (int u = 0; u < n; u++) {
    int l = -1, r = -1, x, y, t;
    int m = int(adj[u].size()) - (u != 0);
    for (int i = 0; i < m; i++) {
      x = adj[u][i];
      if (i == 0) {
        l = st[x];
        r = fi[x];
        t = sz[x];
        continue;
      }
      y = adj[u][i - 1];
      if (sz[x] != sz[y]) {
        g[u].push_back({{l, r}, t});
        l = st[x];
        r = fi[x];
        t = sz[x];
      } else {
        r = fi[x];
      }
    }
    if (l != -1) g[u].push_back({{l, r}, t});
  }
  while (q--) {
    int ty, v;
    scanf("%d %d", &ty, &v);
    v--;
    if (ty == 1) {
      long long int d;
      scanf("%lld", &d);
      d %= mod;
      D[v] = (D[v] + d) % mod;
      long long int k = 1LL * d * sz[v];
      add(0, n, k);
      add(st[v], fi[v], -k);
      for (auto it : g[v]) {
        k = 1LL * d * (n - it.second);
        add(it.first.first, it.first.second, k);
      }
    } else {
      long long int ans = (D[v] + (get(st[v]) % mod * inv) % mod) % mod;
      printf("%lld \n", ans);
    }
  }
  return 0;
}