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 Pow(int a, int b, int md, int res = 1) {
  for (; b; b >>= 1, a = 1ll * a * a % md) {
    if (b & 1) res = 1ll * res * a % md;
  }
  return res;
}
const int N = 1.5e5 + 10, MOD = 998244353, LOG = 21, SQ = 600;
int E[N], second[N], P[LOG][N], H[N], A[N], ps[N], St[N], Ft[N], D[N], n, q,
    tim, invn;
vector<int> adj[N];
vector<pair<int, int> > Q;
void add(int l, int r, int val) {
  ps[l] = (ps[l] + val) % MOD;
  ps[r] = (ps[r] + MOD - val) % MOD;
}
void preDFS(int v, int p = 0) {
  St[v] = tim++;
  second[v] = 1;
  E[St[v]] = v;
  if (!p) P[0][v] = v;
  for (int i = 1; i < LOG; i++) {
    P[i][v] = P[i - 1][P[i - 1][v]];
  }
  for (int u : adj[v]) {
    if (u != p) {
      H[u] = H[v] + 1, P[0][u] = v;
      preDFS(u, v);
      second[v] += second[u];
    }
  }
  Ft[v] = tim;
}
int getpar(int v, int h) {
  for (int i = LOG; i--;)
    if (h & (1 << i)) v = P[i][v];
  return v;
}
void relax() {
  if (!(int)Q.size()) return;
  for (pair<int, int> x : Q) {
    int p = x.first, d = x.second;
    D[p] = (D[p] + d) % MOD;
  }
  for (int v = 1; v <= n; v++) {
    if (!D[v]) continue;
    A[v] = (A[v] + D[v]) % MOD;
    for (int u : adj[v]) {
      if (u == P[0][v]) continue;
      add(St[u], Ft[u], 1ll * D[v] * (n - second[u]) % MOD * invn % MOD);
    }
    add(St[1], St[v], 1ll * D[v] * second[v] % MOD * invn % MOD);
    add(Ft[v], Ft[1], 1ll * D[v] * second[v] % MOD * invn % MOD);
    D[v] = 0;
  }
  for (int i = 0; i < tim; i++) {
    ps[i + 1] = (ps[i + 1] + ps[i]) % MOD;
    A[E[i]] = (A[E[i]] + ps[i]) % MOD;
    ps[i] = 0;
  }
  Q = {};
}
int main() {
  scanf("%d%d", &n, &q);
  invn = Pow(n, MOD - 2, MOD);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  preDFS(1);
  for (; q; q--) {
    if (q % SQ == 0) relax();
    int t, v, d;
    scanf("%d%d", &t, &v);
    if (t == 1) {
      scanf("%d", &d);
      Q.push_back({v, d});
    } else {
      int R = 0;
      for (pair<int, int> x : Q) {
        int p = x.first;
        d = x.second;
        if (St[p] <= St[v] && Ft[v] <= Ft[p]) {
          if (v == p)
            R = (R + d) % MOD;
          else {
            int u = getpar(v, H[v] - H[p] - 1);
            R = (R + 1ll * d * (n - second[u]) % MOD * invn % MOD) % MOD;
          }
        } else
          R = (R + 1ll * d * second[p] % MOD * invn % MOD) % MOD;
      }
      printf("%d\n", (R + A[v]) % MOD);
    }
  }
  return 0;
}