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;
int sz[150005], head[150005];
int par[150005];
long long val[150005];
long long mod = 998244353;
vector<int> vv[150005];
int TT;
int st[150005], en[150005];
long long bigmod(long long n, long long po) {
  if (po == 0) return 1;
  if (po == 1) return n;
  if (po % 2 == 0) {
    long long ret = bigmod(n, po / 2);
    ret *= ret;
    ret %= mod;
    return ret;
  }
  long long ret = bigmod(n, po - 1);
  ret *= n;
  ret %= mod;
  return ret;
}
void dfs1(int u, int p) {
  par[u] = p;
  sz[u] = 1;
  for (int i = 0; i < vv[u].size(); i++) {
    int v = vv[u][i];
    if (v == p) continue;
    dfs1(v, u);
    sz[u] += sz[v];
  }
}
void dfs2(int u, int p, int Last) {
  head[u] = Last;
  int id = 0;
  st[u] = ++TT;
  for (int i = 0; i < vv[u].size(); i++) {
    int v = vv[u][i];
    if (v == p) continue;
    if (id == 0)
      dfs2(v, u, Last);
    else
      dfs2(v, u, v);
    id++;
  }
  en[u] = TT;
}
bool comp(int a, int b) { return sz[a] > sz[b]; }
long long tree[150005 * 4], lazy[150005 * 4];
void Down(int node, int left, int right) {
  if (lazy[node]) {
    tree[left] = (tree[left] + lazy[node]) % mod;
    tree[right] = (tree[right] + lazy[node]) % mod;
    lazy[left] = (lazy[left] + lazy[node]) % mod;
    lazy[right] = (lazy[right] + lazy[node]) % mod;
    lazy[node] = 0;
  }
}
int id = 0;
void update(int node, int b, int e, int l, int r, long long v) {
  if (l > r) return;
  if (b > r || e < l) return;
  if (b >= l && e <= r) {
    tree[node] += v;
    tree[node] %= mod;
    lazy[node] += v;
    lazy[node] %= mod;
    return;
  }
  int left = 2 * node, right = left + 1, mid = (b + e) / 2;
  Down(node, left, right);
  update(left, b, mid, l, r, v);
  update(right, mid + 1, e, l, r, v);
  tree[node] = (tree[left] + tree[right]) % mod;
}
long long query(int node, int b, int e, int i) {
  if (b > i || e < i) return 0;
  if (b == e && b == i) return tree[node];
  int left = 2 * node, right = left + 1, mid = (b + e) / 2;
  Down(node, left, right);
  long long p = query(left, b, mid, i);
  long long q = query(right, mid + 1, e, i);
  long long ret = (p + q) % mod;
  return ret;
}
long long fnc(int u) {
  if (u == 0) return 0;
  int v = head[u];
  int p = par[v];
  long long ret = TT - sz[v];
  ret *= val[p];
  ret %= mod;
  long long zz = fnc(p);
  ret += zz;
  ret %= mod;
  return ret;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  long long inv = bigmod(n, mod - 2);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    vv[u].push_back(v);
    vv[v].push_back(u);
  }
  dfs1(1, 0);
  for (int i = 1; i <= n; i++) sort(vv[i].begin(), vv[i].end(), comp);
  dfs2(1, 0, 1);
  for (int cs = 1; cs <= q; cs++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      id++;
      int v;
      long long d;
      scanf("%d %lld", &v, &d);
      long long z = sz[v];
      z *= d;
      z %= mod;
      update(1, 1, n, 1, st[v] - 1, z);
      update(1, 1, n, en[v] + 1, n, z);
      val[v] += d;
      val[v] %= mod;
      if (vv[v].size() > 1 || v == 1) {
        if (vv[v].size()) {
          int u = vv[v][0];
          if (u == par[v]) u = vv[v][1];
          z = n - sz[u];
          z *= d;
          z %= mod;
          update(1, 1, n, st[u], en[u], z);
        }
      }
    } else {
      int v;
      scanf("%d", &v);
      long long ans = query(1, 1, n, st[v]);
      long long z = fnc(v);
      ans += z;
      ans %= mod;
      ans *= inv;
      ans %= mod;
      ans += val[v];
      ans %= mod;
      printf("%lld\n", ans);
    }
  }
}