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 long long maxn = 150100 + 7;
const long long mod = 998244353;
const long long INF = 1e9 + 7;
const long long mlog = 20;
long long SQ;
long long n, q;
vector<long long> adj[maxn];
vector<pair<long long, long long>> qu;
long long h[maxn], child[maxn];
long long par[mlog][maxn];
long long timer = 0;
long long st[maxn], en[maxn];
long long ex[maxn], sav[maxn], add[maxn];
void dfs(long long v, long long p = n + 1) {
  child[v] = 1;
  st[v] = timer++;
  for (long long u : adj[v]) {
    if (u == p) continue;
    h[u] = h[v] + 1;
    par[0][u] = v;
    dfs(u, v);
    child[v] += child[u];
  }
  en[v] = timer;
}
long long getpar(long long v, long long l) {
  l = h[v] - l;
  for (long long j = 0; j < mlog; j++)
    if (l & (1 << j)) v = par[j][v];
  return 1ll * v;
}
void REBUILD() {
  qu.clear();
  for (long long i = 0; i < n; i++) {
    if (!add[i]) continue;
    long long val = 1ll * add[i] * child[i];
    sav[0] += val;
    sav[st[i]] -= val;
    sav[en[i]] += val;
    sav[n] -= val;
    val = 1ll * add[i] * n;
    sav[st[i]] += val;
    sav[st[i] + 1] -= val;
    for (long long u : adj[i])
      if (h[u] == h[i] + 1) {
        sav[st[u]] += 1ll * add[i] * (n - child[u]);
        sav[en[u]] -= 1ll * add[i] * (n - child[u]);
      }
  }
  for (long long i = 1; i < n; i++) {
    sav[i] += sav[i - 1];
    ex[i] += sav[i];
  }
  ex[0] += sav[0];
  for (long long i = 0; i < n; i++) sav[i] = add[i] = 0;
}
inline long long pw(long long a, long long b) {
  long long res = 1ll;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = a * a % mod;
    b /= 2;
  }
  return res;
}
int32_t main() {
  scanf("%lld%lld", &n, &q);
  SQ = max(1ll, (long long)sqrt(n / log2(n))) * 2;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    u--, v--;
    adj[v].push_back(u), adj[u].push_back(v);
  }
  dfs(0);
  for (long long j = 1; j < mlog; j++)
    for (long long i = 0; i < n; i++) par[j][i] = par[j - 1][par[j - 1][i]];
  long long invn = pw(n, mod - 2);
  for (long long qcnt = 1; qcnt <= q; qcnt++) {
    long long type;
    scanf("%lld", &type);
    if (type == 1) {
      long long v, d;
      scanf("%lld%lld", &v, &d);
      v--;
      qu.push_back({v, d});
      add[v] += d;
    }
    if (type == 2) {
      long long v;
      scanf("%lld", &v);
      v--;
      long long ans = ex[st[v]];
      for (auto qx : qu) {
        long long u = qx.first;
        long long d = qx.second;
        if (st[v] < st[u] || st[v] >= en[u])
          ans += 1ll * child[u] * d;
        else if (v == u)
          ans += 1ll * n * d;
        else {
          long long p = getpar(v, h[u] + 1);
          ans += 1ll * (n - child[p]) * d;
        }
      }
      ans = ((ans % mod) * invn) % mod;
      printf("%lld\n", ans);
    }
    if (qcnt % SQ == 0) REBUILD();
  }
  return 0;
}