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>
const int N = 2e5 + 10, Mod = 998244353;
namespace Fast_IO {
const int bufsize = 1 << 22;
inline char Getchar() {
  static char buf[bufsize], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, bufsize, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <typename T = int>
inline T read() {
  T x = 0, f = 0;
  char ch = Getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = Getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = Getchar();
  return f ? -x : x;
}
char Buf[bufsize], *fout = Buf;
inline void flush() { fwrite(Buf, 1, fout - Buf, stdout), fflush(stdout); }
inline void print(char *s) {
  while (*s) *fout++ = *s++;
}
inline void print(char c) { *fout++ = c; }
template <typename T>
inline void Write(T u) {
  u > 9 ? Write(u / 10), *fout++ = u % 10 + '0' : *fout++ = u + '0';
}
template <typename T>
inline void write(T u, char ch = ' ') {
  if (u < 0) print('-'), u = -u;
  Write(u), print(ch);
}
}  // namespace Fast_IO
using namespace Fast_IO;
int tot = 1, fir[N], nex[N << 1], got[N << 1];
int idx = 0, par[N], dep[N], siz[N], son[N], top[N], dfn[N], tag[N];
inline int Pow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1LL * a * a % Mod)
    if (b & 1) res = 1LL * res * a % Mod;
  return res;
}
inline void AddEdge(int u, int v) {
  nex[++tot] = fir[u], fir[u] = tot, got[tot] = v;
  nex[++tot] = fir[v], fir[v] = tot, got[tot] = u;
}
inline void dfs(int u, int fa) {
  dep[u] = dep[par[u] = fa] + 1, siz[u] = 1;
  for (int i = fir[u]; i; i = nex[i])
    if (got[i] != fa) {
      dfs(got[i], u), siz[u] += siz[got[i]];
      if (siz[got[i]] > siz[son[u]]) son[u] = got[i];
    }
}
inline void Dfs(int u, int fa) {
  dfn[u] = ++idx, top[u] = u == son[fa] ? top[fa] : u;
  if (son[u]) Dfs(son[u], u);
  for (int i = fir[u]; i; i = nex[i])
    if (got[i] != fa && got[i] != son[u]) Dfs(got[i], u);
}
struct BIT {
  int f[N];
  inline void Add(int u, int dat) {
    for (; u < N; u += ((u) & (-u))) (f[u] += dat) %= Mod;
  }
  inline int Ask(int u) {
    int res = 0;
    for (; u; u -= ((u) & (-u))) (res += f[u]) %= Mod;
    return res;
  }
  inline int Ask(int l, int r) { return (Ask(r) - Ask(l - 1) + Mod) % Mod; }
} T;
int main() {
  int n = read(), q = read(), invn = Pow(n, Mod - 2);
  for (int i = 1; i < n; ++i) AddEdge(read(), read());
  dfs(1, 0), Dfs(1, 0);
  for (int Q = 1; Q <= q; ++Q) {
    int opt = read(), u = read();
    if (opt == 1) {
      int d = read();
      T.Add(1, 1LL * d * siz[u] % Mod);
      if (u != 1) T.Add(dfn[u], 1LL * d * (n - siz[u]) % Mod);
      if (son[u]) T.Add(dfn[son[u]], 1LL * d * (Mod - siz[son[u]]) % Mod);
      (tag[u] += d) %= Mod;
    } else {
      int ans = 0;
      for (int fu = top[u]; fu != 1; fu = top[u = par[fu]])
        (ans += T.Ask(dfn[fu], dfn[u])) %= Mod,
            (ans += 1LL * tag[par[fu]] * (Mod - siz[fu]) % Mod) %= Mod;
      write(1LL * (ans + T.Ask(dfn[u])) * invn % Mod, '\n');
    }
  }
  return flush(), 0;
}