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 P = 998244353;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < 48 || c > 57; c = getchar())
    if (!(c ^ 45)) f = 1;
  for (; c >= 48 && c <= 57; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  f ? x = -x : 0;
}
const int N = 150005;
struct edge {
  int to, nxt;
} e[N << 1];
int n, q, et, head[N], sz[N], fa[N], sn[N], dep[N], dt, dfn[N], nfd[N], tp[N],
    T[N], tag[N];
inline int ksm(int x, int q = P - 2) {
  int r = 1;
  for (; q; q >>= 1, x = 1ll * x * x % P)
    if (q & 1) r = 1ll * r * x % P;
  return r;
}
inline void adde(int x, int y) { e[++et] = (edge){y, head[x]}, head[x] = et; }
inline void add(int x, int y) {
  if (x)
    for (; x <= n; x += x & (-x)) (T[x] += y) %= P;
}
inline int qry(int x) {
  int r = 0;
  for (; x; x -= x & (-x)) (r += T[x]) %= P;
  return r;
}
inline void dfs1(int x, int f) {
  dep[x] = dep[f] + 1, fa[x] = f, sz[x] = 1;
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != f) {
      dfs1(e[i].to, x), sz[x] += sz[e[i].to];
      if (sz[e[i].to] >= sz[sn[x]]) sn[x] = e[i].to;
    }
}
inline void dfs2(int x, int top) {
  dfn[x] = ++dt, tp[x] = top;
  if (sn[x]) dfs2(sn[x], top);
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != fa[x] && e[i].to != sn[x]) dfs2(e[i].to, e[i].to);
}
inline void Chang(int x, int w) {
  int gxs = P + 1 - 1ll * ksm(n) * sz[sn[x]] % P,
      gxf = 1ll * ksm(n) * sz[x] % P * w % P;
  (tag[x] += w) %= P, gxs = 1ll * gxs * w % P;
  add(dfn[sn[x]], gxs), add(dfn[sn[x]] + sz[sn[x]], P - gxs), add(dfn[x], w),
      add(dfn[x] + 1, P - w);
  add(1, gxf), add(dfn[x], P - gxf), add(dfn[x] + sz[x], gxf),
      add(n + 1, P - gxf);
}
inline int Query(int x) {
  int res = 0;
  res = qry(dfn[x]);
  while (x > 1)
    x = tp[x],
    (res += tag[fa[x]] * (P + 1 - 1ll * ksm(n) * sz[x] % P) % P) %= P,
    x = fa[x];
  return res;
}
signed main() {
  read(n), read(q);
  for (int i = 1, x, y; i < n; i++) read(x), read(y), adde(x, y), adde(y, x);
  dfs1(1, 0), dfs2(1, 1), memset(T, 0, sizeof(T)), memset(tag, 0, sizeof(tag));
  for (int i = 1, fg, x, y; i <= q; i++) {
    read(fg), read(x);
    if (fg & 1)
      read(y), Chang(x, y);
    else
      printf("%d\n", Query(x));
  }
  return 0;
}