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 = 2e5 + 7;
const int MOD = 998244353;
int sum(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
int sub(int a, int b) { return a - b < 0 ? a - b + MOD : a - b; }
int mul(int a, int b) { return (1LL * a * b) % MOD; }
int power(int a, int b) {
  if (!b) return 1;
  int r = power(a, b / 2);
  r = mul(r, r);
  if (b % 2) r = mul(r, a);
  return r;
}
int tr[4 * MAXN];
int lazy[4 * MAXN];
int combine(int a, int b) { return sum(a, b); }
void propagate(int u, int l, int r) {
  if (lazy[u]) {
    tr[u] = sum(tr[u], lazy[u]);
    if (l != r) {
      lazy[u * 2] = sum(lazy[u * 2], lazy[u]);
      lazy[u * 2 + 1] = sum(lazy[u * 2 + 1], lazy[u]);
    }
    lazy[u] = 0;
  }
}
void update(int u, int l, int r, int x, int y, int v) {
  propagate(u, l, r);
  if (r < x || y < l) return;
  if (x <= l && r <= y) {
    lazy[u] = sum(lazy[u], v);
    propagate(u, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(u * 2, l, mid, x, y, v);
  update(u * 2 + 1, mid + 1, r, x, y, v);
  tr[u] = combine(tr[u * 2], tr[u * 2 + 1]);
}
int query(int u, int l, int r, int x, int y) {
  propagate(u, l, r);
  if (x <= l && r <= y) return tr[u];
  int mid = (l + r) / 2;
  if (y <= mid) return query(u * 2, l, mid, x, y);
  if (mid < x) return query(u * 2 + 1, mid + 1, r, x, y);
  return combine(query(u * 2, l, mid, x, y),
                 query(u * 2 + 1, mid + 1, r, x, y));
}
vector<int> edg[MAXN];
int shuru[MAXN], shesh[MAXN];
int sbtr[MAXN], biggest[MAXN];
int parent[MAXN];
int tym;
void dfs1(int u, int p) {
  sbtr[u] = 1;
  parent[u] = p;
  shuru[u] = ++tym;
  for (int v : edg[u]) {
    if (v == p) continue;
    dfs1(v, u);
    sbtr[u] += sbtr[v];
    if (sbtr[v] > sbtr[biggest[u]]) biggest[u] = v;
  }
  shesh[u] = tym;
}
int checkpoint[MAXN];
int saved[MAXN];
void dfs2(int u) {
  for (int v : edg[u]) {
    if (v == parent[u]) continue;
    if (v == biggest[u]) {
      checkpoint[v] = checkpoint[u];
    } else {
      checkpoint[v] = v;
    }
    dfs2(v);
  }
}
int n;
int invn;
void change(int u, int d) {
  saved[u] = sum(saved[u], d);
  update(1, 1, n, shuru[u], shuru[u], d);
  if (biggest[u] != 0) {
    int v = biggest[u];
    int prob = mul(n - sbtr[v], invn);
    prob = mul(prob, d);
    update(1, 1, n, shuru[v], shesh[v], prob);
  }
  if (sbtr[u] < n) {
    int prob = mul(sbtr[u], invn);
    prob = mul(d, prob);
    update(1, 1, n, 1, n, prob);
    update(1, 1, n, shuru[u], shesh[u], sub(0, prob));
  }
}
int ask(int u) {
  int rt = query(1, 1, n, shuru[u], shuru[u]);
  while (checkpoint[u] > 1) {
    u = checkpoint[u];
    rt = sum(rt, mul(saved[parent[u]], mul(n - sbtr[u], invn)));
    u = parent[u];
  }
  return rt;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> n >> q;
  invn = power(n, MOD - 2);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    edg[u].push_back(v);
    edg[v].push_back(u);
  }
  dfs1(1, 0);
  checkpoint[1] = 1;
  dfs2(1);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int u, d;
      cin >> u >> d;
      change(u, d);
    } else {
      int u;
      cin >> u;
      cout << ask(u) << "\n";
    }
  }
  return 0;
}