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 INF = 2e18;
const int INFi = 2e9;
const int maxN = 2e5 + 1;
const int md = 998244353;
const int md2 = 1e9 + 7;
double getTime() { return clock() / (double)CLOCKS_PER_SEC; };
int mul(int a, int b) { return (1ll * a * b) % md; }
int add(int a, int b) {
  if (a + b >= md) return a + b - md;
  return a + b;
}
int sub(int a, int b) {
  if (a - b < 0) return a - b + md;
  return a - b;
}
int binpow(int a, int b) {
  if (b <= 0) return 1;
  if (b % 2) return mul(a, binpow(a, b - 1));
  int mid = binpow(a, b / 2);
  return mul(mid, mid);
}
int rev(int a) { return binpow(a, md - 2); }
int fenw[maxN];
int q;
void upd(int i, int x) {
  for (; i < maxN; i = i | (i + 1)) fenw[i] = add(fenw[i], x);
}
int get(int i) {
  int res = 0;
  for (; ~i; i = ((i + 1) & i) - 1) res = add(res, fenw[i]);
  return res;
}
int dp[maxN];
vector<pair<int, int>> req1[maxN];
vector<int> req2[maxN];
int ans[maxN];
vector<int> g[maxN];
void dfs(int v, int p) {
  dp[v] = 1;
  for (auto &u : g[v]) {
    if (u == p) continue;
    dfs(u, v);
    dp[v] += dp[u];
  }
}
void dfs_calc(int v, int p) {
  for (auto &req : req1[v]) {
    upd(req.first, mul(req.second, dp[v]));
  }
  for (auto &u : g[v]) {
    if (u == p) continue;
    dfs_calc(u, v);
  }
}
bool cmp(int &a, int &b) { return dp[a] < dp[b]; }
int n;
void re_root(int v, int p) {
  for (auto &req : req1[v]) {
    upd(req.first, mul(req.second, sub(n, dp[v])));
  }
  for (auto &req : req2[v]) {
    ans[req] = mul(get(req), rev(n));
  }
  int prev = n;
  sort((g[v]).begin(), (g[v]).end(), cmp);
  for (auto &u : g[v]) {
    if (u == p) continue;
    int cur = n - dp[u];
    if (cur != prev) {
      for (auto &req : req1[v]) {
        upd(req.first, mul(req.second, sub(cur, prev)));
      }
      prev = cur;
    }
    re_root(u, v);
  }
  for (auto &req : req1[v]) {
    upd(req.first, mul(req.second, sub(dp[v], prev)));
  }
}
void solve() {
  cin >> n >> q;
  for (int _ = 0; (_) < n - 1; ++(_)) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  vector<int> qw;
  for (int i = 0; (i) < q; ++(i)) {
    int t, v;
    cin >> t >> v;
    if (t == 1) {
      int d;
      cin >> d;
      req1[v].emplace_back(i, d);
    } else {
      qw.push_back(i);
      req2[v].push_back(i);
    }
  }
  dfs(1, -1);
  dfs_calc(1, -1);
  re_root(1, -1);
  for (auto &i : qw) cout << ans[i] << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int tests = 1;
  for (int _ = 0; (_) < tests; ++(_)) {
    solve();
  }
  return 0;
}