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, t, a, b, ans, invn;
long long fen[MAXN];
long long lazy[MAXN];
int stt[MAXN], fnt[MAXN];
int timer = 1;
int sz[MAXN];
vector<int> adj[MAXN];
vector<pair<int, int> > child[MAXN];
vector<int> big;
void UPDATE(int pos, long long val) {
  for (; pos < MAXN; pos += pos & -pos) fen[pos] = (fen[pos] + val) % mod;
}
void ADD(int l, int r, long long val) {
  UPDATE(l, +val);
  UPDATE(r, -val);
}
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 par) {
  stt[node] = timer++;
  for (int v : adj[node])
    if (v != par) {
      sz[node] += dfs(v, node), child[node].push_back({fnt[v], v});
    }
  fnt[node] = timer;
  return ++sz[node];
}
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, 1);
  for (int i = 1; i <= n; i++)
    if (child[i].size() > SQ) big.push_back(i);
  invn = powmod(n, mod - 2);
  while (m--) {
    cin >> t;
    if (t == 1) {
      cin >> v >> x;
      ADD(1, n + 1, x * (sz[v] * invn % mod));
      ADD(stt[v], fnt[v], -x * (sz[v] * invn % mod));
      ADD(stt[v] + 1, fnt[v], x * ((n - sz[v] + 1) * invn % mod));
      ADD(stt[v], stt[v] + 1, x);
      if (child[v].size() > SQ) {
        lazy[v] = (lazy[v] + x) % mod;
        continue;
      }
      for (pair<int, int> p : child[v]) {
        int u = p.second;
        ADD(stt[u], fnt[u], x * ((sz[v] - sz[u] - 1) * invn % mod));
      }
    }
    if (t == 2) {
      cin >> v;
      ans = GET(stt[v]);
      for (int u : big)
        if (stt[u] <= stt[v] && fnt[v] <= fnt[u] && u != v) {
          int x = lower_bound(child[u].begin(), child[u].end(),
                              pair<int, int>(fnt[v], 0))
                      ->second;
          ans = (ans + lazy[u] * ((sz[u] - sz[x] - 1) * invn % mod)) % mod;
        }
      if (ans < 0) ans += mod;
      cout << ans << "\n";
    }
  }
  return 0;
}