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 = 150001;
struct Sub {
  int l, r, sz;
};
vector<int> nei[maxN];
vector<Sub> subs[maxN];
int t[4 * maxN];
int sz[maxN], subLt, subl[maxN], subr[maxN];
int cache[maxN];
void Add(int v, int vl, int vr, int l, int r, int val) {
  if (vl == l && vr == r) {
    t[v] += val;
    if (t[v] >= mod) t[v] -= mod;
    return;
  }
  int vm = (vl + vr) / 2;
  if (r <= vm)
    Add(v * 2, vl, vm, l, r, val);
  else if (l > vm)
    Add(v * 2 + 1, vm + 1, vr, l, r, val);
  else {
    Add(v * 2, vl, vm, l, vm, val);
    Add(v * 2 + 1, vm + 1, vr, vm + 1, r, val);
  }
}
int Get(int v, int vl, int vr, int i) {
  if (vl == vr) return t[v];
  int vm = (vl + vr) / 2;
  int res = 0;
  if (i <= vm)
    res = t[v] + Get(v * 2, vl, vm, i);
  else
    res = t[v] + Get(v * 2 + 1, vm + 1, vr, i);
  if (res > mod) res -= mod;
  return res;
}
void InitSz(int v, int par) {
  sz[v] = 1;
  for (auto to : nei[v])
    if (to != par) {
      InitSz(to, v);
      sz[v] += sz[to];
    }
}
void Dfs(int v, int par) {
  subl[v] = ++subLt;
  sort(nei[v].begin(), nei[v].end(),
       [](int u, int v) { return sz[u] < sz[v]; });
  for (int to : nei[v])
    if (to != par) {
      Dfs(to, v);
      if (!subs[v].empty() && sz[to] == subs[v].back().sz)
        subs[v].back().r = subr[to];
      else
        subs[v].push_back({subl[to], subr[to], sz[to]});
    }
  subr[v] = subLt;
}
int BinPow(long long a, int p) {
  if (p == 0) return 1;
  if (p % 2) return (a * BinPow(a, p - 1)) % mod;
  return BinPow((a * a) % mod, p / 2);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n, q;
  cin >> n >> q;
  int nInv = BinPow(n, mod - 2);
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    nei[u].push_back(v);
    nei[v].push_back(u);
  }
  InitSz(1, 0);
  Dfs(1, 0);
  queue<int> upd;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int v, val;
      cin >> v >> val;
      if (cache[v]) {
        cache[v] += val;
        if (cache[v] >= mod) cache[v] -= mod;
      } else {
        cache[v] = val;
        upd.push(v);
      }
    } else {
      while (!upd.empty()) {
        int v = upd.front();
        int val = cache[v];
        cache[v] = 0;
        upd.pop();
        Add(1, 1, n, 1, n, val);
        for (auto sub : subs[v]) {
          long long add = ((long long)sub.sz * nInv) % mod;
          add = (add * val) % mod;
          if (add) Add(1, 1, n, sub.l, sub.r, -add + mod);
        }
        long long upSize = n - sz[v];
        long long add = (upSize * nInv) % mod;
        add = (add * val) % mod;
        if (!add) continue;
        t[1] -= add;
        while (t[1] < 0) t[1] += mod;
        Add(1, 1, n, subl[v], subr[v], add);
      }
      int v;
      cin >> v;
      cout << Get(1, 1, n, subl[v]) << '\n';
    }
  }
}