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 = 150010;
const int mod = 998244353;
const int inv2 = (mod + 1) / 2;
std::vector<int> g[N];
int idx, id[N], c[N], top[N], son[N], sz[N], fa[N], w[N];
inline void update(int x, int y) {
  for (int i = x; i < N; i += i & -i) c[i] = (c[i] + y) % mod;
}
inline int getsum(int x) {
  int res = 0;
  for (int i = x; i; i -= i & -i) res = (res + c[i]) % mod;
  return res;
}
inline void upd(int l, int r, int x) {
  update(l, x);
  update(r + 1, mod - x);
}
void dfs1(int u, int f) {
  son[u] = 0;
  sz[u] = 1;
  for (auto y : g[u])
    if (y != f) {
      fa[y] = u;
      dfs1(y, u);
      sz[u] += sz[y];
      if (sz[y] > sz[son[u]]) son[u] = y;
    }
}
void dfs2(int u, int f) {
  top[u] = f;
  id[u] = ++idx;
  if (son[u]) dfs2(son[u], f);
  for (auto y : g[u])
    if (y != son[u] && y != fa[u]) dfs2(y, y);
}
long long qpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  int inv = qpow(n, mod - 2);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int v, d, tmp;
      scanf("%d%d", &v, &d);
      tmp = (long long)sz[v] * d % mod;
      w[v] = (w[v] + d) % mod;
      update(1, tmp);
      upd(id[v], id[v] + sz[v] - 1, mod - tmp);
      if (!son[v]) continue;
      upd(id[son[v]], id[son[v]] + sz[son[v]] - 1,
          (long long)(n - sz[son[v]]) * d % mod);
    } else {
      int x, res;
      scanf("%d", &x);
      res = (getsum(id[x]) + (long long)w[x] * n % mod) % mod;
      for (x = top[x]; fa[x]; x = top[fa[x]])
        res = (res + (long long)(n - sz[x]) * w[fa[x]] % mod) % mod;
      printf("%d\n", (long long)res * inv % mod);
    }
  }
}