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;
char buf[1 << 20], *_ = buf, *__ = buf;
template <class T>
inline void read(T &x) {
  x = 0;
  char c = (_ == __ && (__ = (_ = buf) + fread(buf, 1, 1 << 20, stdin), _ == __)
                ? EOF
                : *_++);
  while (c < 48 || c > 57)
    c = (_ == __ && (__ = (_ = buf) + fread(buf, 1, 1 << 20, stdin), _ == __)
             ? EOF
             : *_++);
  while (47 < c && c < 58)
    x = (x << 3) + (x << 1) + (c ^ 48),
    c = (_ == __ && (__ = (_ = buf) + fread(buf, 1, 1 << 20, stdin), _ == __)
             ? EOF
             : *_++);
}
template <class T>
inline void read(T &a, T &b) {
  read(a);
  read(b);
}
template <class T>
inline void read(T &a, T &b, T &c) {
  read(a);
  read(b);
  read(c);
}
const long long MAXN = 15e4 + 8, mod = 998244353, inf = 0x3f3f3f3f;
int n;
struct E {
  int y, nt;
} e[MAXN << 1];
int head[MAXN], cnt;
inline void add(int x, int y) {
  e[++cnt].y = y;
  e[cnt].nt = head[x];
  head[x] = cnt;
}
int dep[MAXN], siz[MAXN], fa[MAXN], son[MAXN];
void dfs1(int x) {
  siz[x] = 1, dep[x] = dep[fa[x]] + 1;
  for (int i = head[x]; i; i = e[i].nt) {
    int y = e[i].y;
    if (siz[y]) continue;
    fa[y] = x;
    dfs1(y);
    siz[x] += siz[y];
    if (!son[x] || siz[son[x]] < siz[y]) son[x] = y;
  }
}
int id[MAXN], top[MAXN], idcnt;
long long d[MAXN];
void dfs2(int x, int tp) {
  id[x] = ++idcnt;
  top[x] = tp;
  if (son[x]) dfs2(son[x], tp);
  for (int i = head[x]; i; i = e[i].nt) {
    int y = e[i].y;
    if (id[y]) continue;
    dfs2(y, y);
  }
}
inline void presum_add(int p, long long v) {
  while (p <= n) {
    (d[p] += v) %= mod;
    p += (p & (-p));
  }
}
inline long long ask(int p) {
  long long res = 0;
  while (p) {
    (res += d[p]) %= mod;
    p -= (p & (-p));
  }
  return res;
}
inline void subtree_add(const int x, const long long v) {
  presum_add(id[x], v);
  presum_add(id[x] + siz[x], mod - v);
}
long long qpow(long long x, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * x % mod;
    x = x * x % mod;
    k >>= 1;
  }
  return res;
}
long long lazy[MAXN];
int main() {
  int q;
  read(n, q);
  for (int i = 1, x, y; i < n; ++i) {
    read(x, y);
    add(x, y);
    add(y, x);
  }
  dfs1(1);
  dfs2(1, 1);
  int op, x;
  long long inv = qpow(n, mod - 2), val;
  while (q--) {
    read(op, x);
    if (op == 1) {
      read(val);
      (lazy[x] += val) %= mod;
      subtree_add(x, (n - siz[x]) * val % mod);
      subtree_add(1, siz[x] * val % mod);
      if (!son[x]) continue;
      subtree_add(son[x], mod - siz[son[x]] * val % mod);
    } else {
      long long res = ask(id[x]);
      for (int i = top[x]; fa[i]; i = top[fa[i]]) {
        (res += mod - siz[i] * lazy[fa[i]] % mod) %= mod;
      }
      printf("%I64d\n", res * inv % mod);
    }
  }
  return 0;
}