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 mol = 998244353;
const int maxn = 150000;
long long tree[4 * maxn + 11], lazy[4 * maxn + 11];
long long tag[maxn + 11];
int siz[maxn + 11], f[maxn + 11], id[maxn + 11], top[maxn + 11], son[maxn + 11];
vector<int> edge[maxn + 11];
int tot = 0;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mol;
    a = a * a % mol;
    b >>= 1;
  }
  return ans;
}
void dfs(int x, int fa) {
  siz[x] = 1;
  f[x] = fa;
  for (auto v : edge[x]) {
    if (v == fa) continue;
    dfs(v, x);
    siz[x] += siz[v];
    if (siz[v] > siz[son[x]]) son[x] = v;
  }
}
void dfs2(int x, int t) {
  top[x] = t;
  id[x] = ++tot;
  if (son[x]) dfs2(son[x], t);
  for (auto v : edge[x]) {
    if (v == f[x] || v == son[x]) continue;
    dfs2(v, v);
  }
}
long long add(long long a, long long b) {
  a += b;
  if (a >= mol) a -= mol;
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += mol;
  return a;
}
void push_up(int rt) { tree[rt] = add(tree[rt << 1], tree[rt << 1 | 1]); }
void push_down(int rt, int l, int r) {
  int mid = (l + r) >> 1;
  long long val = lazy[rt];
  lazy[rt] = 0;
  tree[rt << 1] = add(tree[rt << 1], val * (mid - l + 1) % mol);
  lazy[rt << 1] = add(lazy[rt << 1], val);
  tree[rt << 1 | 1] = add(tree[rt << 1 | 1], val * (r - mid) % mol);
  lazy[rt << 1 | 1] = add(lazy[rt << 1 | 1], val);
}
void update(int rt, int l, int r, int al, int ar, long long val) {
  if (l > ar || r < al) return;
  if (l >= al && r <= ar) {
    tree[rt] = add(tree[rt], val * (r - l + 1) % mol);
    lazy[rt] = add(lazy[rt], val);
    return;
  }
  if (lazy[rt]) push_down(rt, l, r);
  int mid = (l + r) >> 1;
  update(rt << 1, l, mid, al, ar, val);
  update(rt << 1 | 1, mid + 1, r, al, ar, val);
  push_up(rt);
}
long long query(int rt, int l, int r, int pos) {
  if (l == r) return tree[rt];
  if (lazy[rt]) push_down(rt, l, r);
  int mid = (l + r) >> 1;
  if (mid >= pos) return query(rt << 1, l, mid, pos);
  return query(rt << 1 | 1, mid + 1, r, pos);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    edge[u].emplace_back(v);
    edge[v].emplace_back(u);
  }
  dfs(1, 0);
  dfs2(1, 1);
  long long all = 0;
  long long inv = qpow(n, mol - 2);
  while (q--) {
    int op, v;
    scanf("%d %d", &op, &v);
    if (op == 1) {
      int d;
      scanf("%d", &d);
      tag[v] = add(tag[v], d);
      all = add(all, inv * d % mol * siz[v] % mol);
      if (son[v])
        update(1, 1, n, id[son[v]], id[son[v]] + siz[son[v]] - 1,
               inv * d % mol * sub(n - siz[v], siz[son[v]]) % mol);
    } else {
      long long ans =
          add(query(1, 1, n, id[v]), inv * tag[v] % mol * (n - siz[v]) % mol);
      while (v) {
        v = top[v];
        ans =
            add(ans, tag[f[v]] * inv % mol * sub(n - siz[v], siz[f[v]]) % mol);
        v = f[v];
      }
      printf("%lld\n", add(ans, all));
    }
  }
}