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;
inline int readint() {
  int x = 0;
  bool f = 0;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') {
    f = 1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f ? -x : x;
}
const int maxn = 150000 + 5;
int n, q;
vector<int> g[maxn];
int fa[maxn], size[maxn], hson[maxn];
void dfs1(int u) {
  size[u] = 1;
  hson[u] = 0;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa[u]) continue;
    fa[v] = u;
    dfs1(v);
    size[u] += size[v];
    if (size[v] > size[hson[u]]) hson[u] = v;
  }
}
int cnt = 0, top[maxn], pos[maxn];
void dfs2(int u) {
  pos[u] = ++cnt;
  if (hson[u]) {
    top[hson[u]] = top[u];
    dfs2(hson[u]);
  }
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa[u] || v == hson[u]) continue;
    top[v] = v;
    dfs2(v);
  }
}
const long long mod = 998244353;
long long c[maxn];
inline int lowbit(int x) { return x & -x; }
void update(int x, long long k) {
  while (x <= n) {
    c[x] = (c[x] + k) % mod;
    x += lowbit(x);
  }
}
long long query(int x) {
  long long s = 0;
  while (x > 0) {
    s = (s + c[x]) % mod;
    x -= lowbit(x);
  }
  return s;
}
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b != 0) {
    if (b % 2 == 1) ans = ans * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return ans;
}
long long invn;
long long tag[maxn];
int main() {
  n = readint();
  q = readint();
  for (int i = 1; i < n; i++) {
    int u, v;
    u = readint();
    v = readint();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  fa[1] = 0;
  size[0] = 0;
  dfs1(1);
  top[1] = 1;
  dfs2(1);
  memset(c, 0, sizeof(c));
  invn = ksm(n, mod - 2);
  memset(tag, 0, sizeof(tag));
  while (q--) {
    int opt = readint();
    if (opt == 1) {
      int v = readint();
      long long d = readint();
      if (v != 1) update(pos[v], d * (n - size[v]) % mod);
      update(1, d * size[v] % mod);
      if (hson[v]) update(pos[v] + 1, d * (mod - size[hson[v]]) % mod);
      tag[v] = (tag[v] + d) % mod;
    } else {
      int v = readint();
      long long ans = 0;
      while (top[v] != 1) {
        ans = (((ans + query(pos[v])) % mod + mod - query(pos[top[v]] - 1)) %
                   mod +
               tag[fa[top[v]]] * (mod - size[top[v]]) % mod) %
              mod;
        v = fa[top[v]];
      }
      ans = (ans + query(pos[v])) % mod;
      printf("%d\n", (int)(ans * invn % mod));
    }
  }
  return 0;
}