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 N = 1.5e5 + 10, MOD = 998244353;
inline int Read(int &r) {
  int b = 1, c = getchar();
  r = 0;
  while (c < '0' || '9' < c) b = c == '-' ? -1 : b, c = getchar();
  while ('0' <= c && c <= '9')
    r = (r << 1) + (r << 3) + (c ^ '0'), c = getchar();
  return r *= b;
}
inline int Add(const int &a, const int &b) {
  return a + b >= MOD ? a + b - MOD : a + b;
}
inline int Sub(const int &a, const int &b) {
  return a - b < 0 ? a - b + MOD : a - b;
}
inline int Mul(const int &a, const int &b) { return int(1ll * a * b % MOD); }
inline int Pow(const int &a, const int &b) {
  return b ? Mul(Pow(Mul(a, a), b >> 1), (b & 1) ? a : 1) : 1;
}
struct GRAPH {
  int head[N], to[N << 1], nxt[N << 1], ncnt;
  inline void AddEdge(const int &u, const int &v) {
    int p = ++ncnt, q = ++ncnt;
    to[p] = v, nxt[p] = head[u], head[u] = p;
    to[q] = u, nxt[q] = head[v], head[v] = q;
  }
  int operator[](const int &u) { return head[u]; }
} Gr;
struct TREEARY {
  int ary[N], siz;
  void Init(const int &n) { siz = n; }
  void Modify(int pos, const int &ad) {
    for (; pos <= siz; pos += ((pos) & (-(pos)))) ary[pos] = Add(ary[pos], ad);
  }
  int Query(int pos) {
    int ret = 0;
    for (; pos; pos -= ((pos) & (-(pos)))) ret = Add(ret, ary[pos]);
    return ret;
  }
  void Modify(const int &le, const int &ri, const int &ad) {
    Modify(le, ad), Modify(ri + 1, Sub(0, ad));
  }
} Se;
int n, m, ndfn;
int dfn[N], siz[N], tpo[N], son[N], anc[N], tot[N], inv[N];
void InitDFS(const int &u, const int &fa) {
  dfn[u] = ++ndfn, siz[u] = 1, anc[u] = fa;
  for (int it = Gr[u]; it; it = Gr.nxt[it]) {
    int v = Gr.to[it];
    if (v == fa) continue;
    InitDFS(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
void TopDFS(const int &u, const int &tp) {
  tpo[u] = tp;
  if (son[u])
    TopDFS(son[u], tp);
  else
    return;
  for (int it = Gr[u]; it; it = Gr.nxt[it]) {
    int v = Gr.to[it];
    if (v == anc[u] || v == son[u]) continue;
    TopDFS(v, v);
  }
}
void Init() {
  inv[0] = 1;
  for (int i = 1; i < N; i++) inv[i] = Mul(inv[i - 1], i);
  inv[N - 1] = Pow(inv[N - 1], MOD - 2);
  for (int i = N - 2; i >= 0; i--) inv[i] = Mul(inv[i + 1], i + 1);
  for (int i = 1, fac = 1; i < N; fac = Mul(fac, i), i++)
    inv[i] = Mul(inv[i], fac);
}
inline int Frac(const int &a, const int &b) { return Mul(a, inv[b]); }
int main() {
  Init();
  Read(n), Read(m);
  for (int i = 1, u, v; i < n; i++) Gr.AddEdge(Read(u), Read(v));
  InitDFS(1, 0), TopDFS(1, 1);
  Se.Init(n);
  while (m--) {
    int cmd, u, ad;
    Read(cmd), Read(u);
    if (cmd == 1) {
      Read(ad);
      int val = Mul(ad, Frac(siz[u], n));
      Se.Modify(1, n, val);
      Se.Modify(dfn[u], dfn[u] + siz[u] - 1, Sub(0, val));
      if (son[u]) {
        int v = son[u];
        val = Mul(ad, Frac(n - siz[v], n));
        Se.Modify(dfn[v], dfn[v] + siz[v] - 1, val);
      }
      tot[u] = Add(tot[u], ad);
    } else {
      int ans = Add(Se.Query(dfn[u]), tot[u]);
      while (u) {
        u = tpo[u];
        if (anc[u]) {
          int v = anc[u];
          ans = Add(ans, Mul(tot[v], Frac(n - siz[u], n)));
          u = v;
        } else
          break;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}