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>
const int MOD = 998244353;
int modpow(int base, int exp) {
  int ac = 1;
  for (; exp; exp >>= 1) {
    if (exp & 1) ac = 1LL * ac * base % MOD;
    base = 1LL * base * base % MOD;
  }
  return ac;
}
int inverse(int x) { return modpow(x, MOD - 2); }
std::vector<int> edges[150005];
int size[150005];
int par[150005];
int head[150005];
int pre[150005];
int post[150005];
int euler;
int delta[150005];
long long ft[150005];
int N, Q;
void update(int i, long long v) {
  for (i++; i <= N; i += (i & -i)) {
    ft[i] += v;
  }
}
void update(int a, int b, long long v) {
  update(a, v);
  update(b, -v);
}
long long query(int i) {
  long long ac = 0;
  for (i++; i > 0; i -= (i & -i)) {
    ac += ft[i];
  }
  return ac;
}
void dfs_size(int node) {
  size[node] = 1;
  for (int child : edges[node]) {
    edges[child].erase(
        std::find(edges[child].begin(), edges[child].end(), node));
    par[child] = node;
    dfs_size(child);
    size[node] += size[child];
  }
}
void dfs_hld(int node) {
  pre[node] = euler++;
  for (int& child : edges[node]) {
    if (size[child] > size[edges[node][0]]) {
      std::swap(child, edges[node][0]);
    }
  }
  for (int child : edges[node]) {
    head[child] = (child == edges[node][0]) ? head[node] : child;
    dfs_hld(child);
  }
  post[node] = euler;
}
int main() {
  scanf("%d %d", &N, &Q);
  int invN = inverse(N);
  for (int i = 0; i < N - 1; i++) {
    int U, V;
    scanf("%d %d", &U, &V);
    U--, V--;
    edges[U].push_back(V);
    edges[V].push_back(U);
  }
  dfs_size(0);
  dfs_hld(0);
  while (Q--) {
    int T;
    scanf("%d", &T);
    if (T == 1) {
      int V, D;
      scanf("%d %d", &V, &D);
      V--;
      update(0, pre[V], 1LL * size[V] * D);
      update(pre[V], post[V], 1LL * N * D);
      update(post[V], N, 1LL * size[V] * D);
      if (edges[V].size()) {
        int U = edges[V][0];
        update(pre[U], post[U], -1LL * size[U] * D);
      }
      delta[V] = (delta[V] + D) % MOD;
    } else {
      int V;
      scanf("%d", &V);
      V--;
      long long ans = query(pre[V]) % MOD;
      for (V = head[V]; V != 0; V = head[par[V]]) {
        ans -= 1LL * size[V] * delta[par[V]];
      }
      printf("%lld\n", (ans % MOD * invN % MOD + MOD) % MOD);
    }
  }
  return 0;
}