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;
constexpr int MAXN = 150000, MOD = 998244353;
vector<int> adj[MAXN + 2];
int sz[MAXN + 2], lft[MAXN + 2], rht[MAXN + 2], seg[MAXN * 2 + 2],
    light[MAXN + 2], par[MAXN + 2], cnt[MAXN + 2];
int dfs1(int u, int w) {
  par[u] = w;
  sz[u] = 1;
  int mxsz = 0, mxi, n = adj[u].size();
  for (int i = 0; i < n;) {
    if (adj[u][i] == w) {
      adj[u][i] = adj[u][--n];
      adj[u].pop_back();
      continue;
    }
    int csz = dfs1(adj[u][i], u);
    sz[u] += csz;
    if (mxsz < csz) {
      mxsz = csz;
      mxi = i;
    }
    i++;
  }
  if (n > 1) swap(adj[u][0], adj[u][mxi]);
  return sz[u];
}
int nxt = 0;
void dfs2(int u, int w) {
  light[u] = w;
  lft[u] = nxt++;
  int n = adj[u].size();
  if (n) {
    dfs2(adj[u][0], w);
    for (int i = 1; i < n; i++) {
      int v = adj[u][i];
      dfs2(v, v);
    }
  }
  rht[u] = nxt;
}
int reduce(int x) { return x >= MOD ? x - MOD : x; }
int n;
void update(int l, int r, int v) {
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      seg[l] = reduce(seg[l] + v);
      l++;
    }
    if (r & 1) {
      --r;
      seg[r] = reduce(seg[r] + v);
    }
  }
}
int query(int i) {
  i += n;
  long long res = 0;
  do res += seg[i];
  while (i >>= 1);
  return res % MOD;
}
int mpow(int b, int e) {
  int res = 1;
  while (e) {
    if (e & 1) res = (long long)res * b % MOD;
    b = (long long)b * b % MOD;
    e >>= 1;
  }
  return res;
}
int main() {
  int q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  int invn = mpow(n, MOD - 2);
  do {
    char op;
    int v;
    cin >> op >> v;
    if (op == '1') {
      int d;
      cin >> d;
      int val = (long long)d * sz[v] % MOD;
      update(lft[1], rht[1], val);
      update(lft[v], rht[v], MOD - val);
      int i = lft[v] + n;
      seg[i] = (seg[i] + (long long)d * n) % MOD;
      if (adj[v].size()) {
        int w = adj[v][0];
        update(lft[w], rht[w], (long long)d * (n - sz[w]) % MOD);
      }
      cnt[v] = reduce(cnt[v] + d);
    } else {
      int res = query(lft[v]);
      while (light[v]) {
        v = light[v];
        int w = par[v];
        res = (res + (long long)(n - sz[v]) * cnt[w]) % MOD;
        v = w;
      }
      res = (long long)res * invn % MOD;
      cout << res << '\n';
    }
  } while (--q);
}