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;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
int mod = 998244353;
inline int mul(int x, int y) { return 1ll * x * y % mod; }
int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
int sub(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
int sq(int x) { return 1ll * x * x % mod; }
int pow(int a, int b) {
  return b == 0 ? 1 : (b & 1 ? mul(a, sq(pow(a, b / 2))) : sq(pow(a, b / 2)));
}
int inv(int x) { return pow(x, mod - 2); }
const int maxn = 150050;
struct fenwick_tree {
  int c[maxn];
  inline void tadd(int x, int y) {
    while (x <= 150000) {
      c[x] = add(c[x], y);
      x += x & -x;
    }
  }
  inline int sum(int x) {
    int res = 0;
    while (x > 0) {
      res = add(res, c[x]);
      x -= x & -x;
    }
    return res;
  }
  inline void vadd(int l, int r, int v) {
    tadd(l, v);
    tadd(r + 1, mod - v);
  }
} A;
int tim = 0;
int n, q, d[150050], sz[150050], mson[150050], fa[150050], top[150050],
    dfn[150050], accu[150050], u, v;
vector<int> G[150050];
void dfs1(int num) {
  sz[num] = 1;
  mson[num] = 0;
  for (auto ct : G[num]) {
    if (ct == fa[num]) continue;
    fa[ct] = num;
    dfs1(ct);
    sz[num] += sz[ct];
    if (sz[ct] > sz[mson[num]]) mson[num] = ct;
  }
}
void dfs2(int num) {
  tim++;
  dfn[num] = tim;
  if (num == mson[fa[num]])
    top[num] = top[fa[num]];
  else
    top[num] = num;
  if (mson[num] == 0) return;
  dfs2(mson[num]);
  for (auto ct : G[num]) {
    if (ct == mson[num] || ct == fa[num]) continue;
    dfs2(ct);
  }
}
int main() {
  read(n);
  read(q);
  int invn = inv(n);
  for (int i = 1; i < n; i++) {
    read(u);
    read(v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs1(1);
  dfs2(1);
  for (int i = 0; i < q; i++) {
    int typ, v, d;
    read(typ);
    if (typ == 1) {
      read(v);
      read(d);
      A.vadd(1, n, mul(d, mul(sz[v], invn)));
      A.vadd(dfn[v], dfn[v] + sz[v] - 1, mod - mul(d, mul(sz[v], invn)));
      if (mson[v]) {
        A.vadd(dfn[mson[v]], dfn[mson[v]] + sz[mson[v]] - 1,
               mul(d, mul(n - sz[mson[v]], invn)));
      }
      accu[v] = add(accu[v], d);
    } else {
      read(v);
      int cv = add(A.sum(dfn[v]), accu[v]);
      while (1) {
        v = top[v];
        if (v == 1) break;
        cv = add(cv, mul(accu[fa[v]], mul(n - sz[v], invn)));
        v = fa[v];
      }
      cout << cv << "\n";
    }
  }
  return 0;
}