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 = 15e4;
const long long MOD = 998244353;
int N, Q, deg[MAXN + 10];
vector<int> adj[MAXN + 10];
int S[MAXN + 10], E[MAXN + 10], par[MAXN + 10], sz[MAXN + 10], head[MAXN + 10],
    heavy[MAXN + 10], cnt;
long long memo[MAXN + 10];
long long mypow(long long x, long long y) {
  if (y == 0) return 1;
  if (y % 2) return mypow(x, y - 1) * x % MOD;
  long long ret = mypow(x, y / 2);
  return ret * ret % MOD;
}
struct BIT {
  long long tree[MAXN + 10];
  void update(int i, long long x) {
    for (; i <= N; i += (i & -i)) tree[i] = (tree[i] + x) % MOD;
  }
  void update(int l, int r, long long x) {
    update(l, x);
    update(r + 1, (MOD - x) % MOD);
  }
  long long query(int i) {
    long long ret = 0;
    for (; i > 0; i -= (i & -i)) ret = (ret + tree[i]) % MOD;
    return ret;
  }
} bit;
void dfs(int now, int bef) {
  par[now] = bef;
  sz[now] = 1;
  for (int nxt : adj[now]) {
    if (nxt == bef) continue;
    dfs(nxt, now);
    sz[now] += sz[nxt];
  }
}
void hld(int now, int bef) {
  S[now] = E[now] = ++cnt;
  heavy[now] = 0;
  for (int nxt : adj[now])
    if (nxt != bef && sz[heavy[now]] < sz[nxt]) heavy[now] = nxt;
  if (!heavy[now]) return;
  head[heavy[now]] = head[now];
  hld(heavy[now], now);
  for (int nxt : adj[now]) {
    if (nxt == bef || nxt == heavy[now]) continue;
    head[nxt] = nxt;
    hld(nxt, now);
  }
  for (int nxt : adj[now])
    if (nxt != bef) E[now] = max(E[now], E[nxt]);
}
int main() {
  int i, j;
  scanf("%d%d", &N, &Q);
  for (i = 1; i < N; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    deg[u]++;
    deg[v]++;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  head[1] = 1;
  hld(1, 0);
  while (Q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int v;
      long long d;
      scanf("%d%lld", &v, &d);
      if (heavy[v])
        bit.update(S[heavy[v]], E[heavy[v]], d * (N - sz[heavy[v]]) % MOD);
      bit.update(1, N, d * sz[v] % MOD);
      bit.update(S[v], E[v], (MOD - d * sz[v] % MOD) % MOD);
      bit.update(S[v], S[v], d * N % MOD);
      memo[v] += d;
      memo[v] %= MOD;
    } else {
      int v;
      scanf("%d", &v);
      long long ans = bit.query(S[v]);
      while (head[v] != 1) {
        int h = head[v];
        ans += memo[par[h]] * (N - sz[h]);
        ans %= MOD;
        v = par[h];
      }
      long long p = ans, q = N;
      p *= mypow(q, MOD - 2);
      p %= MOD;
      printf("%lld\n", p);
    }
  }
}