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 MX = 2e5 + 7;
const int mod = 998244353;
const double pi = 3.1415926535897932384;
double isp = 1e-13;
using namespace std;
long long qpow(long long a, long long b, long long MOD = mod) {
  for (long long ans = 1;; a = a * a % MOD, b >>= 1) {
    if (b & 1) ans = ans * a % MOD;
    if (!b) return ans;
  }
}
long long inv(long long a, long long MOD = mod) {
  return qpow(a, MOD - 2, MOD);
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long ret = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return ret;
}
long long getInv(int a, int mod) {
  long long x, y;
  long long d = exgcd(a, mod, x, y);
  return d == 1 ? (x % mod + mod) % mod : -1;
}
vector<int> G[MX];
long long dfn[MX], hson[MX], fa[MX], sz[MX], cnt, top[MX], po[MX];
long long c[MX], in;
int n, q;
void upd(int pos, int x) {
  for (; pos <= n; pos += pos & -pos) {
    (c[pos] += x) % mod;
  }
}
long long get_sum(int pos) {
  long long ans = 0;
  while (pos) {
    ans += c[pos];
    pos -= pos & -pos;
  }
  return ans;
}
void dfs1(int x, int f) {
  sz[x] = 1;
  for (int y : G[x]) {
    if (y == f) continue;
    fa[y] = x;
    dfs1(y, x);
    sz[x] += sz[y];
    if (sz[hson[x]] < sz[y]) hson[x] = y;
  }
}
void dfs2(int x, int f) {
  dfn[x] = ++cnt;
  top[x] = f;
  if (!hson[x]) return;
  dfs2(hson[x], f);
  for (int y : G[x]) {
    if (y != fa[x] && y != hson[x]) dfs2(y, y);
  }
}
long long query(int x) {
  long long ans = 0;
  while (top[x] != 1) {
    ans = (ans + get_sum(dfn[x]) - get_sum(dfn[top[x]] - 1) +
           1ll * po[fa[top[x]]] * (mod - sz[top[x]]) % mod) %
          mod;
    ans = (ans + mod) % mod;
    x = fa[top[x]];
  }
  ans += get_sum(dfn[x]);
  ans %= mod;
  return ans * in % mod;
}
void update(int v, long long d) {
  if (v != 1) upd(dfn[v], 1ll * d * (n - sz[v]) % mod);
  upd(1, 1ll * d * sz[v] % mod);
  if (hson[v]) upd(dfn[hson[v]], 1ll * d * (mod - sz[hson[v]]) % mod);
  po[v] += d;
  po[v] %= mod;
  return;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> q;
  in = inv(n);
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  dfs1(1, -1);
  dfs2(1, 1);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int v, d;
      cin >> v >> d;
      update(v, d);
    } else {
      int v;
      cin >> v;
      cout << query(v) << '\n';
    }
  }
}