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("Ofast,no-stack-protector,unroll-loops,fast-math")
using namespace std;
const int N = 150000 + 10;
const long long MOD = 998244353;
const long long INF = 1000000010;
const long long LOG = 18;
const int SQ = 550;
vector<int> H, G[N];
int n, q, fen[N], par[N][LOG], sz[N], st[N], ft[N], Time, sm[N], h[N];
void DFS(int v = 1, int p = 0) {
  par[v][0] = p;
  sz[v] = 1;
  st[v] = ++Time;
  for (int i = 1; i < LOG; i++) par[v][i] = par[par[v][i - 1]][i - 1];
  for (auto u : G[v]) {
    if (u == p) continue;
    h[u] = h[v] + 1;
    DFS(u, v);
    sz[v] += sz[u];
  }
  ft[v] = Time + 1;
}
inline int minu(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
inline void ADD(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
inline int POW(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = res * 1ll * a % MOD;
    b >>= 1;
    a = a * 1ll * a % MOD;
  }
  return res;
}
inline int getpar(int v, int k) {
  for (int i = 0; i < LOG; i++)
    if (k & (1 << i)) v = par[v][i];
  return v;
}
inline void add(int id, int x) {
  for (; id < N; id += id & (-id)) ADD(fen[id], x);
}
inline int get(int id) {
  int res = 0;
  for (; id > 0; id -= id & (-id)) ADD(res, fen[id]);
  return res;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    G[v].push_back(u), G[u].push_back(v);
  }
  for (int i = 1; i <= n; i++) {
    if (G[i].size() > SQ) H.push_back(i);
  }
  DFS();
  for (int i = 1; i <= q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int v, d;
      cin >> v >> d;
      if (G[v].size() > SQ) {
        ADD(sm[v], d);
      } else {
        add(st[v], d * 1ll * n % MOD);
        add(st[v] + 1, minu(0, d * 1ll * n % MOD));
        for (auto u : G[v]) {
          if (u == par[v][0]) {
            int num = d * 1ll * sz[v] % MOD;
            add(1, num);
            add(st[v], minu(0, num));
            add(ft[v], num);
            continue;
          }
          int SZ = n - sz[u];
          add(st[u], d * 1ll * SZ % MOD);
          add(ft[u], minu(0, d * 1ll * SZ % MOD));
        }
      }
    } else {
      int v;
      cin >> v;
      int res = get(st[v]);
      for (auto u : H) {
        if (st[u] <= st[v] && ft[v] <= ft[u]) {
          int vv = getpar(v, h[v] - h[u] - 1);
          int SZ = n - sz[vv];
          ADD(res, SZ * 1ll * sm[u] % MOD);
          continue;
        }
        ADD(res, sz[u] * 1ll * sm[u] % MOD);
      }
      cout << res * 1ll * POW(n, MOD - 2) % MOD << '\n';
    }
  }
  return 0;
}