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>
#pragma GCC optimize("O2")
using namespace std;
const long long MXN = 15e4 + 10;
const long long Mod = 998244353;
const long long SQR = 1000;
long long power(long long a, long long b) {
  return (!b ? 1 : power(a * a % Mod, b / 2) * (b & 1LL ? a : 1) % Mod);
}
long long inv(long long x) { return power(x, Mod - 2); }
long long n, q, invn, timer;
long long Par[MXN], sub[MXN], hvs[MXN], Fen[MXN];
long long hd[MXN], Stm[MXN], Ftm[MXN], Lazy[MXN], Ex[MXN];
vector<long long> adj[MXN];
void Upd(long long p, long long x) {
  for (; p < MXN; p += p & (-p)) Fen[p] = (Fen[p] + x) % Mod;
}
void Upd(long long l, long long r, long long x) { Upd(l, x), Upd(r + 1, -x); }
long long Get(long long p) {
  long long niw = 0;
  for (; p; p -= p & (-p)) niw += Fen[p];
  return niw % Mod;
}
void prep(long long u, long long par) {
  sub[u]++, Par[u] = par;
  for (auto v : adj[u]) {
    if (v == par) continue;
    prep(v, u);
    sub[u] += sub[v];
    if (sub[v] > sub[hvs[u]]) hvs[u] = v;
  }
}
void dfs(long long u, long long par, long long head) {
  Stm[u] = ++timer;
  hd[u] = head;
  if (hvs[u]) dfs(hvs[u], u, head);
  for (auto v : adj[u]) {
    if (v == par || v == hvs[u]) continue;
    dfs(v, u, v);
  }
  Ftm[u] = timer;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q, invn = inv(n);
  for (int i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v), adj[v].push_back(u);
  }
  prep(1, 0);
  dfs(1, 0, 1);
  while (q--) {
    long long typ, u, d;
    cin >> typ >> u;
    if (typ == 1) {
      cin >> d;
      Upd(Stm[1], Stm[u] - 1, d * sub[u] % Mod * invn % Mod);
      Upd(Ftm[u] + 1, Ftm[1], d * sub[u] % Mod * invn % Mod);
      Lazy[u] = (Lazy[u] + d) % Mod;
      if (hvs[u]) {
        long long v = hvs[u];
        Upd(Stm[v], Ftm[v], d * (n - sub[v]) % Mod * invn % Mod);
      }
      Ex[u] += d;
    } else {
      long long ans = Get(Stm[u]), now;
      long long v = u;
      while (v) {
        long long c = hd[v], r = Par[c];
        now = Lazy[r] * (n - sub[c]) % Mod * invn % Mod;
        ans = (ans + now) % Mod;
        v = r;
      }
      ans += Ex[u];
      cout << (ans + Mod) % Mod << '\n';
    }
  }
  return 0;
}