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 mod = 998244353;
const int MAXN = 150010;
const int SQ = 95;
long long n, m, k, u, v, x, y, z, t, a, b, ans, invn;
long long fen[MAXN];
long long lazy[MAXN];
int st[MAXN], en[MAXN];
int timer = 1;
int sz[MAXN], son[MAXN], par[MAXN], h[MAXN], top[MAXN];
vector<int> adj[MAXN];
vector<pair<int, int> > child[MAXN];
vector<int> big;
void UPDATE(int pos, long long val) {
  for (; pos <= n; pos += pos & -pos) fen[pos] = (fen[pos] + val) % mod;
}
long long GET(int pos) {
  long long res = 0;
  for (; pos; pos -= pos & -pos) res += fen[pos];
  return res % mod;
}
long long powmod(long long a, long long b) {
  if (!b) return 1;
  if (b & 1) return a * powmod(a * a % mod, b >> 1) % mod;
  return powmod(a * a % mod, b >> 1);
}
int dfs(int node, int pars) {
  st[node] = timer++;
  for (int v : adj[node]) {
    if (v == pars) continue;
    par[v] = node;
    h[v] = h[node] + 1;
    sz[node] += dfs(v, node);
    son[node] = ((!son[node] || sz[son[node]] < sz[v]) ? v : son[node]);
  }
  en[node] = timer;
  return ++sz[node];
}
void p_dfs(int v, int p) {
  top[v] = p;
  if (sz[v] == 1) return;
  p_dfs(son[v], p);
  for (int u : adj[v])
    if (u != par[v] && u != son[v]) p_dfs(u, u);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(), cout.tie();
  ;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  p_dfs(1, 1);
  invn = powmod(n, mod - 2);
  while (m--) {
    cin >> t;
    if (t == 1) {
      cin >> x >> y;
      z = 1ll * sz[x] * y % mod;
      lazy[x] = (lazy[x] + y) % mod;
      UPDATE(1, z);
      UPDATE(st[x], -z);
      UPDATE(en[x], z);
      if (!son[x]) continue;
      z = 1ll * (n - sz[son[x]]) * y % mod;
      UPDATE(st[son[x]], z);
      UPDATE(en[son[x]], -z);
    }
    if (t == 2) {
      cin >> x;
      ans = GET(st[x]);
      ans = (ans + 1ll * lazy[x] * n) % mod;
      for (int i = top[x]; par[i]; i = top[par[i]])
        ans = (ans + 1ll * (n - sz[i]) * lazy[par[i]]) % mod;
      ans = 1ll * ans * invn % mod;
      if (ans < 0) ans += mod;
      cout << ans << "\n";
    }
  }
  return 0;
}