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 N = 150000 + 10;
const int M = 998244353;
const double eps = 1e-9;
const double PI = acos(-1);
const int oo = 1000000000;
int n, q, L[N], R[N], P[N], T, val[N], fen[N], invN, sz[N], bigC[N];
vector<int> g[N];
pair<int, int> up[N];
int mul(int x, int y) { return 1ll * x * y % M; }
int add(int x, int y) {
  int ret = x + y;
  if (ret > M) ret -= M;
  if (ret < 0) ret += M;
  return ret;
}
void upd(int u, int d) {
  for (; u; u -= u & -u) fen[u] = add(fen[u], d);
}
void upd(int l, int r, int d) {
  upd(r, d);
  upd(l - 1, -d);
}
int get(int u) {
  int ret = 0;
  for (; u <= n; u += u & -u) ret = add(ret, fen[u]);
  return ret;
}
void dfs(int u, int p) {
  P[u] = p;
  L[u] = ++T;
  sz[u] = 1;
  for (int d : g[u]) {
    if (d - p) {
      dfs(d, u);
      sz[u] += sz[d];
    }
  }
  R[u] = T;
}
void dfs(int u, int p, int ls, int cc) {
  int mx = -1, id = -1;
  up[u] = {ls, cc};
  for (int d : g[u]) {
    if (d - p && sz[d] > mx) {
      mx = sz[d];
      id = d;
    }
  }
  bigC[u] = id;
  for (int i = 0; i < g[u].size(); ++i) {
    int d = g[u][i];
    if (d == p) continue;
    if (d - id)
      dfs(d, u, u, i);
    else
      dfs(d, u, ls, cc);
  }
}
int pw(int a, int b) {
  if (!b) return 1;
  int d = pw(a, b / 2);
  d = mul(d, d);
  if (b & 1) d = mul(d, a);
  return d;
}
int main() {
  memset(up, -1, sizeof(up));
  cin >> n >> q;
  invN = pw(n, M - 2);
  for (int a, b, i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1);
  dfs(0, -1, -1, -1);
  for (int t, u, d, i = 0; i < q; ++i) {
    scanf("%d%d", &t, &u);
    --u;
    if (t == 1) {
      scanf("%d", &d);
      val[u] = add(val[u], d);
      int big = bigC[u];
      if (big + 1) upd(L[big], R[big], mul(d, mul(n - sz[big], invN)));
      if (L[u] > 1) upd(1, L[u] - 1, mul(d, mul(sz[u], invN)));
      if (R[u] < n) upd(R[u] + 1, n, mul(d, mul(sz[u], invN)));
    } else {
      int ans = add(val[u], get(L[u]));
      pair<int, int> cur = up[u];
      while (cur.first + 1) {
        int pp = cur.first, ch = g[pp][cur.second];
        ans = add(ans, mul(val[pp], mul(n - sz[ch], invN)));
        cur = up[pp];
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}