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;
bool big[150010];
long long n, q, din[150010], dout[150010], dv, sonsz[150010],
    aff[150010 / 700 + 2][150010], seg[700], fre[150010], par[150010],
    vv[150010], cnt = 0;
vector<long long> vt[150010], bg;
long long ksm(long long x, long long y) {
  if (y == 0) return 1;
  long long ret = ksm(x, y >> 1);
  ret = (ret * ret) % 998244353;
  if (y % 2 == 1) return (ret * x) % 998244353;
  return ret;
}
void dfs(long long x, long long lst) {
  din[x] = ++cnt;
  par[x] = lst;
  long long i;
  for (i = 0; i < vt[x].size(); i++) {
    if (vt[x][i] != lst) dfs(vt[x][i], x);
  }
  dout[x] = cnt;
  sonsz[x] = dout[x] - din[x] + 1;
  return;
}
void prebg() {
  long long i, j, k, x, y;
  for (i = 0; i < bg.size(); i++) {
    x = bg[i];
    for (j = 0; j < vt[x].size(); j++) {
      y = vt[x][j];
      if (par[y] != x) continue;
      for (k = din[y]; k <= dout[y]; k++)
        aff[i][k] = (dv * (n - sonsz[y])) % 998244353;
    }
    for (j = 1; j < din[x]; j++) aff[i][j] = (dv * sonsz[x]) % 998244353;
    for (j = dout[x] + 1; j <= cnt; j++)
      aff[i][j] = (dv * sonsz[x]) % 998244353;
    aff[i][din[x]] = (dv * n) % 998244353;
  }
  return;
}
void cge(long long x, long long val) {
  seg[x / 700] = (seg[x / 700] + val) % 998244353;
  fre[x] = (fre[x] + val) % 998244353;
  return;
}
void update(long long l, long long r, long long val) {
  cge(l, val);
  cge(r + 1, 998244353 - val);
}
long long getans(long long x) {
  long long i, ret = 0;
  for (i = 0; i < x / 700; i++) ret = (ret + seg[i]) % 998244353;
  for (i = (x / 700) * 700; i <= x; i++) ret = (ret + fre[i]) % 998244353;
  return ret;
}
void upd(long long x, long long val) {
  if (big[x]) {
    vv[x] = (vv[x] + val) % 998244353;
    return;
  }
  update(din[x], din[x], val);
  update(1, din[x] - 1, (((val * sonsz[x]) % 998244353) * dv) % 998244353);
  update(dout[x] + 1, cnt, (((val * sonsz[x]) % 998244353) * dv) % 998244353);
  long long i;
  for (i = 0; i < vt[x].size(); i++) {
    if (par[vt[x][i]] != x) continue;
    update(din[vt[x][i]], dout[vt[x][i]],
           (((val * (n - sonsz[vt[x][i]])) % 998244353) * dv) % 998244353);
  }
  return;
}
long long ga(long long x) {
  long long i, ret = 0;
  for (i = 0; i < bg.size(); i++)
    ret = (ret + (aff[i][din[x]] * vv[bg[i]]) % 998244353) % 998244353;
  ret = (ret + getans(din[x])) % 998244353;
  return ret;
}
int main() {
  long long i, x, y, ty;
  scanf("%lld%lld", &n, &q);
  dv = ksm(n, 998244353 - 2);
  for (i = 1; i < n; i++) {
    scanf("%lld%lld", &x, &y);
    x--;
    y--;
    vt[x].push_back(y);
    vt[y].push_back(x);
  }
  dfs(0, -1);
  for (i = 0; i < n; i++)
    if (vt[i].size() >= 700) {
      big[i] = true;
      bg.push_back(i);
    }
  prebg();
  while (q--) {
    scanf("%lld%lld", &ty, &x);
    x--;
    if (ty == 1) {
      scanf("%lld", &y);
      upd(x, y);
    } else
      printf("%lld\n", ga(x));
  }
  return 0;
}