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;
int hed[150005], nex[300005], to[300005];
int pos[150005], top[150005];
int siz[150005], son[150005];
int fa[150005], dep[150005];
int mark[150005];
int cnt, S, n, pn, Q, ans, pans;
int ksm(int x, int num) {
  if (num == 1) return x;
  int j = ksm(x, num / 2);
  if (num & 1)
    return 1ll * j * j % mod * x % mod;
  else
    return 1ll * j * j % mod;
}
void add(int x, int y) {
  cnt++;
  nex[cnt] = hed[x];
  hed[x] = cnt;
  to[cnt] = y;
}
void dfs1(int x, int fi) {
  fa[x] = fi;
  siz[x] = 1;
  dep[x] = dep[fi] + 1;
  for (int i = hed[x]; i != 0; i = nex[i])
    if (to[i] != fa[x]) {
      dfs1(to[i], x);
      siz[x] += siz[to[i]];
      if (siz[to[i]] > siz[son[x]]) son[x] = to[i];
    }
}
void dfs2(int x, int tp) {
  top[x] = tp;
  pos[x] = ++S;
  if (son[x] != 0) dfs2(son[x], tp);
  for (int i = hed[x]; i != 0; i = nex[i])
    if (to[i] != son[x] && to[i] != fa[x]) dfs2(to[i], to[i]);
}
struct tree {
  int f[150005];
  void Ad(int w, int x) {
    while (w <= n) {
      f[w] = ((f[w] + x) % mod + mod) % mod;
      w += w & (-w);
    }
  }
  int ch(int w) {
    int j = 0;
    while (w > 0) {
      j = (j + f[w]) % mod;
      w -= w & (-w);
    }
    return j;
  }
} T;
void solve(int x) {
  x = top[x];
  while (fa[x] != 0) {
    pans = (pans - 1ll * mark[fa[x]] * siz[x] % mod + mod) % mod;
    x = fa[x];
    x = top[x];
  }
}
int main() {
  int opt, x, y;
  scanf("%d%d", &n, &Q);
  pn = ksm(n, mod - 2);
  for (int i = 2; i <= n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  while (Q--) {
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d", &x, &y);
      if (y == 0) continue;
      mark[x] = (mark[x] + y) % mod;
      ans = (ans + 1ll * siz[x] * y % mod) % mod;
      if (son[x] != 0) {
        T.Ad(pos[son[x]], -1ll * siz[son[x]] * y % mod);
        T.Ad(pos[son[x]] + siz[son[x]], 1ll * siz[son[x]] * y % mod);
      }
      T.Ad(pos[x], 1ll * (n - siz[x]) * y % mod);
      T.Ad(pos[x] + siz[x], 1ll * (siz[x] - n) * y % mod);
    } else {
      scanf("%d", &x);
      pans = ans;
      pans = (pans + T.ch(pos[x])) % mod;
      solve(x);
      pans = 1ll * pans * pn % mod;
      printf("%d\n", pans);
    }
  }
  return 0;
}