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 M = (119 << 23) + 1;
const int MX = 150000 + 1;
const int MAG = 400;
const int LOG_N = 19;
int fen_tree[MX], n;
int time_in[MX], time_out[MX], t;
int sz[MX], par[LOG_N][MX], log_n;
vector<int> adj[MX];
map<int, int> heavy;
int add(int a, int b) { return (a += b) < M ? a : a - M; }
int mul(int a, int b) { return add(1LL * a * b % M, M); }
int mod_pow(int a, int b) {
  int res = 1;
  while (b) {
    if (b % 2) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b /= 2;
  }
  return res;
}
void upd(int a, int val) {
  for (; a <= n; a += a & -a) {
    fen_tree[a] = add(fen_tree[a], val);
  }
}
int qry(int a) {
  int res = 0;
  for (; a; a -= a & -a) {
    res = add(res, fen_tree[a]);
  }
  return res;
}
void dfs(int u) {
  if (int(adj[u].size()) > MAG) {
    heavy[u] = 0;
  }
  sz[u] = 1;
  time_in[u] = ++t;
  for (auto c : adj[u]) {
    if (c != par[0][u]) {
      par[0][c] = u;
      dfs(c);
      sz[u] += sz[c];
    }
  }
  time_out[u] = t;
}
bool is_anc(int u, int v) {
  return time_in[u] <= time_in[v] && time_out[v] <= time_out[u];
}
int suc(int u, int v) {
  for (int i = log_n - 1; i >= 0; --i) {
    if (!is_anc(par[i][v], u)) {
      v = par[i][v];
    }
  }
  return v;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int m;
  cin >> n >> m;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  time_out[0] = t;
  for (log_n = 1; (1 << log_n) < 2 * n; ++log_n) {
    for (int j = 1; j <= n; ++j) {
      par[log_n][j] = par[log_n - 1][par[log_n - 1][j]];
    }
  }
  while (m--) {
    int type;
    cin >> type;
    if (type == 1) {
      int v, d;
      cin >> v >> d;
      int base = mul(d, mod_pow(n, M - 2));
      upd(1, mul(sz[v], base));
      upd(time_in[v], mul(n - sz[v], base));
      upd(time_out[v] + 1, mul(sz[v] - n, base));
      if (heavy.count(v)) {
        heavy[v] = add(heavy[v], base);
      } else {
        for (auto c : adj[v]) {
          if (c != par[0][v]) {
            upd(time_in[c], mul(-sz[c], base));
            upd(time_out[c] + 1, mul(sz[c], base));
          }
        }
      }
    } else {
      int v;
      cin >> v;
      int ans = qry(time_in[v]);
      for (auto [u, val] : heavy) {
        if (u != v && is_anc(u, v)) {
          ans = add(ans, mul(val, -sz[suc(u, v)]));
        }
      }
      cout << ans << "\n";
    }
  }
}