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 mod = 998244353;
inline void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
struct bit {
  int n;
  vector<int> b;
  bit(int n) : n(n), b(n + 1) {}
  void update(int l, int r, int x) {
    for (; l <= n; l += l & -l) add(b[l], x);
    for (++r; r <= n; r += r & -r) add(b[r], mod - x);
  }
  int query(int p) {
    int x = 0;
    for (; p > 0; p -= p & -p) add(x, b[p]);
    return x;
  }
};
int Pow(int a, int b) {
  int x = 1;
  for (; b > 0; b >>= 1) {
    if (b & 1) x = (long long)x * a % mod;
    a = (long long)a * a % mod;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<vector<int>> adj(n);
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    adj[--u].push_back(--v);
    adj[v].push_back(u);
  }
  vector<int> size(n), heavy(n, -1), Q(1, 0), parent(n, -1), head(n);
  for (int i = 0, u; i < n; ++i) {
    size[u = Q[i]] = 1;
    for (auto v : adj[u])
      if (parent[u] != v) Q.push_back(v), parent[v] = u;
  }
  for (int i = n - 1, u; i >= 0; --i) {
    u = Q[i];
    for (auto v : adj[u])
      if (parent[u] != v) {
        size[u] += size[v];
        if (heavy[u] == -1 || size[v] > size[heavy[u]]) heavy[u] = v;
      }
  }
  for (int u = 0; u < n; ++u)
    if (u == 0 || heavy[parent[u]] != u)
      for (int v = u; v != -1; v = heavy[v]) head[v] = u;
  int t = 0;
  vector<int> dt(n), ft(n);
  function<void(int, int)> dfs = [&](int u, int p) {
    dt[u] = ++t;
    for (auto v : adj[u])
      if (v != p) dfs(v, u);
    ft[u] = t;
  };
  dfs(0, -1);
  const int inv = Pow(n, mod - 2);
  bit b(n);
  vector<int> sum(n);
  for (int type, v, d; q--;) {
    cin >> type >> v, --v;
    if (type == 1) {
      cin >> d;
      const int mul = (long long)inv * d % mod;
      b.update(dt[v], dt[v], d);
      b.update(dt[0], ft[0], (long long)size[v] * mul % mod);
      b.update(dt[v], ft[v], mod - (long long)size[v] * mul % mod);
      int h = heavy[v];
      if (h != -1) b.update(dt[h], ft[h], (long long)(n - size[h]) * mul % mod);
      add(sum[v], mul);
      continue;
    }
    int ans = b.query(dt[v]);
    while (true) {
      v = head[v];
      if (parent[v] == -1) break;
      add(ans, (long long)(n - size[v]) * sum[parent[v]] % mod);
      v = parent[v];
    }
    cout << ans << "\n";
  }
  return 0;
}