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;
inline void read(int &x) {
  int flg = 1;
  char ch;
  while (!isdigit(ch = getchar()))
    if (ch == '-') flg = -flg;
  for (x = ch - '0'; isdigit(ch = getchar()); x = x * 10 + ch - '0')
    ;
  x *= flg;
}
const int MAXN = 150050;
const int mod = 998244353;
int n, m, fir[MAXN], to[MAXN << 1], nxt[MAXN << 1], cnt;
void link(int u, int v) {
  to[++cnt] = v;
  nxt[cnt] = fir[u];
  fir[u] = cnt;
  to[++cnt] = u;
  nxt[cnt] = fir[v];
  fir[v] = cnt;
}
int sz[MAXN], fa[MAXN], top[MAXN], dep[MAXN], son[MAXN], dfn[MAXN], tmr;
void dfs1(int u, int ff) {
  dep[u] = dep[fa[u] = ff] + (sz[u] = 1);
  for (int i = fir[u], v; i; i = nxt[i])
    if ((v = to[i]) != ff) {
      dfs1(v, u);
      sz[u] += sz[v];
      if (sz[v] > sz[son[u]]) son[u] = v;
    }
}
void dfs2(int u, int tp) {
  top[u] = tp;
  dfn[u] = ++tmr;
  if (son[u]) dfs2(son[u], tp);
  for (int i = fir[u], v; i; i = nxt[i])
    if ((v = to[i]) != fa[u] && v != son[u]) dfs2(v, v);
}
int qpow(int a, int b) {
  int re = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) re = 1ll * re * a % mod;
  return re;
}
int ans[MAXN], T[MAXN];
void upd(int x, int v) {
  while (x <= n) (T[x] += v) %= mod, x += x & -x;
}
int qsum(int x) {
  int re = 0;
  while (x) (re += T[x]) %= mod, x -= x & -x;
  return re;
}
void add(int l, int r, int v) { upd(l, v), upd(r + 1, mod - v); }
int Query(int x) {
  int re = 0;
  while (x) {
    x = top[x];
    (re += 1ll * (n - sz[x]) * ans[fa[x]] % mod) %= mod;
    x = fa[x];
  }
  return re;
}
int main() {
  read(n), read(m);
  for (int i = 1, x, y; i < n; ++i) read(x), read(y), link(x, y);
  dfs1(1, 0);
  dfs2(1, 1);
  int invn = qpow(n, mod - 2), op, v, d;
  while (m--) {
    read(op), read(v);
    if (op == 1) {
      read(d);
      (ans[v] += d) %= mod;
      if (son[v])
        add(dfn[son[v]], dfn[son[v]] + sz[son[v]] - 1,
            1ll * (n - sz[son[v]]) * d % mod);
      if (dfn[v] > 1) add(1, dfn[v] - 1, 1ll * sz[v] * d % mod);
      if (dfn[v] + sz[v] <= n) add(dfn[v] + sz[v], n, 1ll * sz[v] * d % mod);
    } else {
      int Ans = 1ll * (Query(v) + qsum(dfn[v])) * invn % mod;
      (Ans += ans[v]) %= mod;
      printf("%d\n", Ans);
    }
  }
}