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;
int N, M, q, x, y, K, T, d;
int poz[150100], v[150100], nr[150100], lvl[150100], p[150100], cmp[150100],
    viz[150100];
vector<int> g[150100], c[150100];
int val[150100];
int tbd[150100];
int pw(int x, int y) {
  if (!y) return 1;
  if (y % 2) return (1LL * x * pw(x, y - 1)) % 998244353;
  int z = pw(x, y / 2);
  return (1LL * z * z) % 998244353;
}
vector<int> AIB[150100];
inline int zeros(int x) { return x & (-x); }
inline void add(int C, int x, int q) {
  ++x;
  for (int i = x; i <= c[C].size(); i += zeros(i)) {
    AIB[C][i] += q;
    if (AIB[C][i] >= 998244353)
      AIB[C][i] -= 998244353;
    else if (AIB[C][i] < 0)
      AIB[C][i] += 998244353;
  }
}
inline int calc(int C, int x) {
  ++x;
  int ret = 0;
  for (int i = x; i > 0; i -= zeros(i)) {
    ret += AIB[C][i];
    if (ret >= 998244353)
      ret -= 998244353;
    else if (ret < 0)
      ret += 998244353;
  }
  return ret;
}
void dfs(int x) {
  viz[x] = 1;
  nr[x] = 1;
  int ind = -1, nrc = -1;
  for (auto y : g[x]) {
    if (viz[y]) continue;
    lvl[y] = lvl[x] + 1;
    p[y] = x;
    dfs(y);
    if (nr[y] > nrc) {
      ind = y;
      nrc = nr[y];
    }
    nr[x] += nr[y];
  }
  if (nrc == -1) {
    vector<int> C;
    C.push_back(x);
    ++K;
    c[K] = C;
    cmp[x] = K;
  } else {
    c[cmp[ind]].push_back(x);
    cmp[x] = cmp[ind];
  }
}
void heavy_path(int r) {
  lvl[r] = 1;
  dfs(r);
  for (int i = 1; i <= K; ++i) {
    reverse(c[i].begin(), c[i].end());
    for (int j = 0; j < c[i].size(); ++j) poz[c[i][j]] = j;
  }
  for (int i = 1; i <= K; ++i) AIB[i].resize(c[i].size() + 2, 0);
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> q;
  int iN = pw(N, 998244353 - 2);
  for (int i = 1; i < N; ++i) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  heavy_path(1);
  int total = 0;
  for (int i = 1; i <= q; ++i) {
    cin >> T;
    if (T == 1) {
      cin >> x >> d;
      total = (total + 1LL * d * nr[x]) % 998244353;
      val[x] = (val[x] + 1LL * d * (N - nr[x])) % 998244353;
      tbd[x] = (tbd[x] + d) % 998244353;
      int P = poz[x], C = cmp[x];
      if (P < c[C].size() - 1) {
        int y = c[C][P + 1];
        long long meh = (N + 998244353 - nr[x] - nr[y]) % 998244353;
        meh = (meh * d) % 998244353;
        add(C, P + 1, meh);
      }
    } else {
      cin >> x;
      long long meh = calc(cmp[x], poz[x]);
      int root = c[cmp[x]][0];
      while (root != 1) {
        int bottom = p[root];
        meh = (1LL * tbd[bottom] * (N - nr[bottom] - nr[root] + 998244353) +
               meh) %
              998244353;
        meh += calc(cmp[bottom], poz[bottom]);
        if (meh >= 998244353) meh -= 998244353;
        root = c[cmp[bottom]][0];
      }
      cout << ((1LL * (total + val[x] + meh) * iN) % 998244353) << "\n";
    }
  }
}