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 maxn = 300030, mod = 998244353;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 0;
  while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
int n, q, el, head[maxn], to[maxn], nxt[maxn], b[maxn], dep[maxn], sz[maxn],
    son[maxn], fa[maxn], top[maxn], lft[maxn], rig[maxn], cnt, tag[maxn];
inline void add(int u, int v) {
  to[++el] = v;
  nxt[el] = head[u];
  head[u] = el;
}
inline int qpow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
inline void update(int p, int v) {
  for (int i = p; i <= n; i += i & -i) b[i] = (b[i] + v) % mod;
}
inline void update(int l, int r, int v) {
  update(l, v);
  update(r + 1, (mod - v) % mod);
}
inline int query(int p) {
  int s = 0;
  for (int i = p; i; i -= i & -i) s = (s + b[i]) % mod;
  return s;
}
void dfs1(int u, int f) {
  dep[u] = dep[fa[u] = f] + 1;
  sz[u] = 1;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == f) continue;
    dfs1(v, u);
    sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int topf) {
  top[u] = topf;
  lft[u] = ++cnt;
  if (son[u]) dfs2(son[u], topf);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa[u] || v == son[u]) continue;
    dfs2(v, v);
  }
  rig[u] = cnt;
}
void update_tree(int u, int d) {
  tag[u] = (tag[u] + d) % mod;
  if (son[u])
    update(lft[son[u]], rig[son[u]], 1ll * (n - sz[son[u]]) * d % mod);
  if (u != 1) {
    update(1, lft[u] - 1, 1ll * sz[u] * d % mod);
    if (rig[u] != n) update(rig[u] + 1, n, 1ll * sz[u] * d % mod);
  }
}
int query_tree(int u) {
  int s = (1ll * n * tag[u] + query(lft[u])) % mod;
  while (u) {
    s = (s + 1ll * tag[fa[top[u]]] * (n - sz[top[u]])) % mod;
    u = fa[top[u]];
  }
  return s;
}
int main() {
  n = read();
  q = read();
  int inv = qpow(n, mod - 2);
  for (int i = (1); i <= (n - 1); i++) {
    int u = read(), v = read();
    add(u, v);
    add(v, u);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  while (q--) {
    int tp = read(), u = read();
    if (tp == 1)
      update_tree(u, 1ll * inv * read() % mod);
    else
      printf("%d\n", query_tree(u));
  }
}