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>
struct {
  inline operator int() {
    int x;
    return scanf("%d", &x), x;
  }
} read;
const int maxn = 150005, mod = 998244353;
long long bit[maxn];
std::vector<int> G[maxn];
int top[maxn], deep[maxn], fa[maxn], size[maxn], son[maxn];
long long tag[maxn];
inline long long power(long long x, int k) {
  if (k < 0) k += mod - 1;
  long long res = 1;
  while (k) {
    if (k & 1) (res *= x) %= mod;
    (x *= x) %= mod;
    k >>= 1;
  }
  return res;
}
void modify(int l, int r, long long x) {
  for (int k = r; k; k -= k & -k) bit[k] += x;
  for (int k = l - 1; k; k -= k & -k) bit[k] -= x;
}
long long query(int p) {
  long long res = 0;
  for (int k = p; k < maxn; k += k & -k) res += bit[k];
  return res % mod;
}
void dfs1(int u) {
  deep[u] = deep[fa[u]] + 1;
  size[u] = 1;
  for (int v : G[u])
    if (v != fa[u]) {
      fa[v] = u;
      dfs1(v);
      size[u] += size[v];
      if (size[v] > size[son[u]]) son[u] = v;
    }
}
int id[maxn], iid[maxn], ip;
void dfs2(int u) {
  if (!top[u]) top[u] = u;
  id[u] = ++ip;
  if (son[u]) {
    top[son[u]] = top[u];
    dfs2(son[u]);
    for (int v : G[u])
      if (v != fa[u] and v != son[u]) dfs2(v);
  }
}
int main() {
  int n = read, q = read;
  for (int i = 1; i < n; i++) {
    int u = read, v = read;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs1(1);
  dfs2(1);
  long long inv = power(n, -1);
  while (q--) {
    int o = read;
    if (o == 1) {
      int u = read, s = son[u];
      long long x = read * inv % mod;
      modify(1, n, x * n % mod);
      if (s) modify(id[s], id[s] + size[s] - 1, size[s] * (mod - x) % mod);
      modify(1, n, (n - size[u]) * (mod - x) % mod);
      modify(id[u], id[u] + size[u] - 1, (n - size[u]) * x % mod);
      tag[u] += x;
    }
    if (o == 2) {
      int u = read;
      long long ans = query(id[u]);
      while (u) {
        u = top[u];
        (ans += size[u] * (mod - tag[fa[u]])) %= mod;
        u = fa[u];
      }
      printf("%lld\n", ans);
    }
  }
}