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 = 1.5e5 + 5;
const long long MOD = 998244353;
long long mod(long long a, long long b) { return ((a % b) + b) % b; }
long long extended_euclid(long long a, long long b, long long &x,
                          long long &y) {
  long long xx = y = 0;
  long long yy = x = 1;
  while (b) {
    long long q = a / b;
    long long t = b;
    b = a % b;
    a = t;
    t = xx;
    xx = x - q * xx;
    x = t;
    t = yy;
    yy = y - q * yy;
    y = t;
  }
  return a;
}
long long mod_inverse(long long a, long long n) {
  long long x, y;
  long long d = extended_euclid(a, n, x, y);
  if (d > 1) return -1;
  return mod(x, n);
}
long long N, Q;
vector<int> G[n];
long long P[n], S[n];
int L[n], R[n];
int CC = 0;
int changed[n];
vector<int> X;
int XV[n], Z[n];
long long U[n];
tuple<int, int, int> Qs[n];
int DFS(int v, int p) {
  P[v] = p;
  L[v] = CC++;
  R[v] = L[v];
  S[v] = 1;
  for (int c : G[v]) {
    if (c != p) {
      R[v] = DFS(c, v);
      S[v] += S[c];
    }
  }
  return R[v];
}
void DFS3(int v, long long z) {
  z = z + U[v];
  XV[v] = (XV[v] + z) % MOD;
  for (int c : G[v]) DFS3(c, z);
}
int DFS2(int v, long long z) {
  long long up = 0;
  for (int c : G[v]) {
    assert(c != P[v]);
    int r = DFS2(c, (z + Z[v] * (N - S[c])) % MOD);
    U[c] = (r + Z[c] * S[c]) % MOD;
    up += U[c];
  }
  XV[v] = (XV[v] + z + up + Z[v] * N) % MOD;
  for (int c : G[v]) {
    U[c] = (up - U[c]) % MOD;
  }
  return up % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> Q;
  long long DIVN = mod_inverse(N, MOD);
  for (int i = 0; i < N; ++i) changed[i] = -1;
  for (int i = 0; i < N - 1; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  DFS(0, -1);
  for (int i = 1; i < N; ++i) {
    G[i].erase(remove(G[i].begin(), G[i].end(), P[i]), G[i].end());
  }
  int q1 = 0, q2 = 0;
  for (int q = 0; q < Q; q++) {
    int t, v, d;
    cin >> t;
    if (t == 1) {
      cin >> v >> d;
      v--;
      q1++;
    } else {
      cin >> v;
      v--;
      q2++;
    }
    Qs[q] = {t, v, d};
  }
  long long cc = 0;
  float fac = max(0.1, min(15.0, 6.0 * q1 / q2));
  cerr << "q1, q2: " << q1 << " " << q2 << endl;
  cerr << "fac: " << fac << endl;
  for (int q = 0; q < Q; q++) {
    int t, v, d;
    tie(t, v, d) = Qs[q];
    if (t == 1) {
      long long c = (d * DIVN) % MOD;
      Z[v] = (Z[v] + c) % MOD;
      if (changed[v] == -1) {
        X.push_back(v);
        changed[v] = cc++;
      }
    } else {
      if (cc * cc > fac * Q) {
        DFS2(0, 0);
        DFS3(0, 0);
        for (int x : X) changed[x] = -1;
        X.clear();
        fill_n(Z, N, 0);
        cc = 0;
      }
      long long s = XV[v];
      for (int x : X) {
        if (x == v) {
          s = (s + Z[x] * N) % MOD;
        } else if (L[v] < L[x] || L[v] > R[x]) {
          s = (s + Z[x] * S[x]) % MOD;
        } else {
          int l = 0, r = G[x].size() - 1, m;
          while (l < r) {
            m = (l + r + 1) / 2;
            if (L[G[x][m]] > L[v]) {
              r = m - 1;
            } else {
              l = m;
            }
          }
          s = (s + Z[x] * (N - S[G[x][l]])) % MOD;
        }
      }
      cout << s << "\n";
    }
  }
  return 0;
}