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 long long int N = 150000 + 10;
const long long int LOG = 20;
const long long int SQR = 400;
const int MOD = 998244353;
const long long int INF = 1e18 + 10;
int n, q;
int sz[N], L[N], R[N], _time;
vector<int> adj[N];
vector<pair<int, int> > vc[N], g[N];
long long int fen[N], all[N];
inline void go(int v = 1, int p = 0) {
  sz[v] = 1;
  for (int u : adj[v]) {
    if (u == p) continue;
    go(u, v);
    sz[v] += sz[u];
  }
  for (int u : adj[v]) {
    if (u == p) continue;
    vc[v].push_back({sz[u], u});
  }
  int p1 = 0, p2 = vc[v].size() - 1;
  sort(vc[v].begin(), vc[v].end());
  while (0 <= p2) {
    int cnst = vc[v][p2].first, cn = 0;
    while (0 <= p2 && vc[v][p2].first == cnst) {
      adj[v][p1] = vc[v][p2].second;
      cn++;
      p1++;
      p2--;
    }
    g[v].push_back({cnst, cn});
  }
  if (v - 1) adj[v].pop_back();
  return;
}
inline void dfs(int v = 1) {
  L[v] = _time++;
  for (int u : adj[v]) dfs(u);
  R[v] = _time;
  return;
}
inline void add(int pos, long long int val) {
  for (pos++; pos <= n + 1; pos += (pos & (-pos))) {
    fen[pos] += val;
  }
  return;
}
inline long long int get(int pos) {
  long long int res = 0;
  for (pos++; pos; pos -= (pos & (-pos))) {
    res += fen[pos];
  }
  return res;
}
inline long long int inv(long long int x, int t = MOD - 2) {
  if (t == 0) return 1ll;
  return ((t & 1) ? x : 1ll) * inv(x * x % MOD, t >> 1) % MOD;
}
int main() {
  scanf("%d%d", &n, &q);
  long long int invN = inv(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  go();
  dfs();
  while (q--) {
    int type, v;
    long long int d;
    scanf("%d%d", &type, &v);
    if (type == 1) {
      scanf("%lld", &d);
      all[v] += d;
      long long int val = sz[v] * d % MOD;
      add(0, val);
      add(L[v], -val);
      add(R[v], val);
      int l = L[v] + 1, r;
      for (auto it : g[v]) {
        r = l + it.first * it.second;
        val = (n - it.first) * d % MOD;
        add(l, val);
        add(r, -val);
        l = r;
      }
    } else {
      all[v] %= MOD;
      printf("%lld\n",
             ((all[v] + invN * (get(L[v]) % MOD) % MOD) % MOD + MOD) % MOD);
    }
  }
  return 0;
}