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 mod = 998244353;
const long long INF = 0x3f3f3f3f;
const long double EPS = 1e-6;
long long qread() {
  char c = getchar();
  long long neg = 1;
  long long val = 0;
  while (c != '-' && !isdigit(c)) c = getchar();
  if (c == '-') {
    c = getchar();
    neg = -1;
  }
  while (isdigit(c)) {
    val = (val << 3) + (val << 1) + c - '0';
    c = getchar();
  }
  return neg * val;
}
const long long N = 200020;
long long n, m;
struct Fenwick {
  long long arr[N + 15];
  void modify(long long x, long long d) {
    for (; x <= N; x += x & (-x)) {
      arr[x] = (arr[x] + d) % mod;
    }
  }
  long long query(long long x) {
    long long res = 0;
    for (; x; x -= x & (-x)) {
      res = (res + arr[x]) % mod;
    }
    return res;
  }
} fenwick;
long long top[N], dfn[N], dep[N], fa[N], hson[N], sz[N];
vector<long long> g[N];
void dfs1(long long u) {
  sz[u] = 1;
  for (auto v : g[u]) {
    if (v == fa[u]) continue;
    dep[v] = dep[u] + 1;
    fa[v] = u;
    dfs1(v);
    sz[u] += sz[v];
    if (sz[v] > sz[hson[u]]) hson[u] = v;
  }
}
long long in[N], out[N];
long long tot;
void dfs2(long long u, long long t) {
  top[u] = t;
  in[u] = dfn[u] = ++tot;
  if (hson[u]) dfs2(hson[u], t);
  for (auto v : g[u])
    if (v != hson[u] && v != fa[u]) dfs2(v, v);
  out[u] = tot;
}
long long qpow(long long a, long long b = mod - 2) {
  long long ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
  }
  return ans % mod;
}
long long lazy[N];
void solve() {
  n = qread();
  m = qread();
  for (long long i = 0; i < (long long)n - 1; i++) {
    long long u = qread(), v = qread();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  fa[1] = 1;
  dep[1] = 1;
  dfs1(1);
  dfs2(1, 1);
  long long invn = qpow(n);
  while (m--) {
    long long op = qread();
    if (op == 1) {
      long long u = qread(), d = qread();
      if (u != 1) fenwick.modify(dfn[u], d * (n - sz[u]) % mod);
      fenwick.modify(1, d * sz[u] % mod);
      if (hson[u]) fenwick.modify(dfn[u] + 1, d * (mod - sz[hson[u]]) % mod);
      lazy[u] = (lazy[u] + d) % mod;
    } else {
      long long u = qread();
      long long ans = 0;
      while (top[u] != 1) {
        ans = (((ans + fenwick.query(dfn[u])) % mod + mod * 2 -
                fenwick.query(dfn[top[u]] - 1)) %
                   mod +
               lazy[fa[top[u]]] * (mod * 2 - sz[top[u]]) % mod) %
              mod;
        u = fa[top[u]];
      }
      ans = (ans + fenwick.query(dfn[u])) % mod;
      printf("%lld\n", (long long)(ans * invn % mod));
    }
  }
}
signed main() {
  long long T = 1;
  while (T--) solve();
  return 0;
}