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 inf = 1e9;
const long long inf_ll = 1e18;
struct ancestor {
  vector<int> vin, vout, d, sz;
  vector<vector<int>>&adj, p;
  int n, t = 0, e;
  void dfs(int i, int v) {
    vin[i] = t++;
    p[i][0] = v, d[i] = d[v] + 1;
    for (int j = 1; j <= e; j++) p[i][j] = p[p[i][j - 1]][j - 1];
    for (auto& j : adj[i])
      if (j != v) dfs(j, i), sz[i] += sz[j];
    vout[i] = t - 1;
  }
  ancestor(vector<vector<int>>& _adj, int root) : adj(_adj), n(_adj.size()) {
    vin.resize(n), vout.resize(n);
    d.assign(n, 0), sz.assign(n, 1);
    e = ceil(log2(n));
    p.assign(n, vector<int>(e + 1));
    dfs(root, root);
  }
  bool is_ancestor(int i, int j) const {
    return vin[i] <= vin[j] && vout[i] >= vout[j];
  }
  int ktha(int i, int k) const {
    for (int j = 0; (1 << j) <= k; j++)
      if (k & (1 << j)) i = p[i][j];
    return i;
  }
  int lca(int i, int j) const {
    if (is_ancestor(i, j)) return i;
    if (is_ancestor(j, i)) return j;
    for (int k = e; k >= 0; k--)
      if (!is_ancestor(p[i][k], j)) i = p[i][k];
    return p[i][0];
  }
  int step(int i, int j) const {
    return is_ancestor(i, j) ? ktha(j, d[j] - d[i] - 1) : p[i][0];
  }
};
const long long mod = 998244353;
struct segtree {
  vector<int> a, b;
  vector<long long> tree, u;
  int n;
  segtree(int _n) {
    n = 1 << int(ceil(log2(_n)));
    tree.assign(2 * n, 0);
    a.resize(2 * n);
    b.resize(2 * n);
    u.assign(2 * n, 0);
    for (int i = 2 * n - 1; i > 0; i--) {
      if (i >= n) {
        a[i] = b[i] = i - n;
      } else {
        a[i] = a[2 * i];
        b[i] = b[2 * i + 1];
        tree[i] = (tree[2 * i] + tree[2 * i + 1]) % mod;
      }
    }
  }
  void push(int i) {
    if (u[i] == 0) return;
    tree[i] = (tree[i] + u[i]) % mod;
    if (i < n) {
      u[2 * i] = (u[2 * i] + u[i]) % mod;
      u[2 * i + 1] = (u[2 * i + 1] + u[i]) % mod;
    }
    u[i] = 0;
  }
  void update(int l, int r, long long x, int i = 1) {
    if (l <= a[i] && r >= b[i]) u[i] = (u[i] + x) % mod;
    push(i);
    if (l > b[i] || r < a[i] || l <= a[i] && r >= b[i]) return;
    update(l, r, x, 2 * i);
    update(l, r, x, 2 * i + 1);
    tree[i] = (tree[2 * i] + tree[2 * i + 1]) % mod;
  }
  long long query(int l, int r, int i = 1) {
    push(i);
    if (l <= a[i] && r >= b[i]) return tree[i];
    if (l > b[i] || r < a[i]) return 0;
    return (query(l, r, 2 * i) + query(l, r, 2 * i + 1)) % mod;
  }
};
template <typename T>
struct fenwick {
  int length;
  vector<T> tree;
  fenwick(int size) {
    length = size + 1;
    tree.assign(length + 1, 0);
  }
  fenwick(vector<T>& v) {
    length = v.size() + 1;
    tree.assign(length + 1, 0);
    for (int i = 0; i < v.size(); i++) update(i, v[i]);
  }
  void update(int i, T val) {
    i++;
    while (i < length) {
      tree[i] = (tree[i] + val) % mod;
      i += i & -i;
    }
  }
  T prefix(int i) {
    i++;
    T out = 0;
    while (i > 0) {
      out = (out + tree[i]) % mod;
      ;
      i -= i & -i;
    }
    return out;
  }
};
template <typename T>
T inv(T a, T m) {
  T r = 1;
  for (T k = m - 2; k; k >>= 1) {
    if (k & 1) r = (r * a) % m;
    a = (a * a) % m;
  }
  return r % m;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q, b;
  cin >> n >> q;
  b = sqrt(n);
  vector<vector<int>> adj(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v), adj[v].push_back(u);
  }
  ancestor a(adj, 0);
  vector<int> h;
  vector<long long> c(n);
  for (int i = 0; i < n; i++)
    if (adj[i].size() > b) h.push_back(i);
  long long ninv = inv((long long)n, mod), g = 0;
  fenwick<long long> f(n + 100);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long v, d;
      cin >> v >> d;
      v--;
      g = (g + d * a.sz[v] % mod * ninv % mod) % mod;
      long long u = d * (n - a.sz[v]) % mod * ninv % mod;
      f.update(a.vin[v], u);
      f.update(a.vout[v] + 1, mod - u);
      if (adj[v].size() > b)
        c[v] = (c[v] + d) % mod;
      else {
        for (auto& j : adj[v])
          if (a.vin[j] > a.vin[v]) {
            u = mod - d * a.sz[j] % mod * ninv % mod;
            f.update(a.vin[j], u);
            f.update(a.vout[j] + 1, mod - u);
          }
      }
    } else {
      long long v;
      cin >> v;
      v--;
      long long x = (f.prefix(a.vin[v]) + g) % mod;
      for (auto& i : h)
        if (a.is_ancestor(i, v) && i != v) {
          x = (x + mod - c[i] * a.sz[a.step(i, v)] % mod * ninv % mod) % mod;
        }
      cout << x << "\n";
    }
  }
}