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>
#pragma GCC optimize("Ofast,unroll-loops")
using namespace std;
const int maxn = 15e4 + 7, blo = 255, mo = 998244353;
int n, q, cl = 0, p[maxn], sz[maxn], lb[maxn], rb[maxn];
int_fast64_t inv, one = 0, rec[maxn], dat[maxn];
vector<int> vt, g[maxn];
vector<vector<int>> hsz;
inline int qp(int_fast64_t x, int t = mo - 2) {
  int_fast64_t re = 1;
  while (t) {
    if (t & 1) re = re * x % mo;
    x = x * x % mo, t >>= 1;
  }
  return re;
}
inline void add(int i, int x) {
  for (; i <= cl + 1; i += i & -i) {
    dat[i] += x;
    if (dat[i] >= mo) dat[i] -= mo;
  }
}
inline int sum(int i) {
  int_fast64_t re = 0;
  for (; i > 0; i -= i & -i) {
    re += dat[i];
  }
  re += one;
  return re % mo;
}
void dfs(int u, int f) {
  lb[u] = cl++;
  sz[u] = 1;
  p[u] = f;
  for (register int i = (0); i < (g[u].size()); ++i) {
    int& v = g[u][i];
    if (v == f) continue;
    dfs(v, u);
    sz[u] += sz[v];
  }
  rb[u] = cl;
}
void gsz(int u, int f, int k) {
  hsz.back()[u] = k;
  for (register int i = (0); i < (g[u].size()); ++i) {
    int& v = g[u][i];
    if (v == f) continue;
    gsz(v, u, k);
  }
}
int main() {
  cin.tie(0), cout.tie(0), ios_base::sync_with_stdio(false);
  cin >> n >> q;
  inv = qp(n);
  for (register int i = (1); i < (n); ++i) {
    int u, v;
    cin >> u >> v, --u, --v;
    g[u].emplace_back(v), g[v].emplace_back(u);
  }
  dfs(0, -1);
  for (register int i = (0); i < (n); ++i) {
    if (g[i].size() > blo) {
      vt.emplace_back(i);
      hsz.emplace_back(*new vector<int>);
      hsz.back().reserve(n);
      int s = 0;
      for (register int j = (0); j < (g[i].size()); ++j) {
        int& v = g[i][j];
        if (v == p[i]) continue;
        gsz(v, i, sz[v]);
        s += sz[v];
      }
      if (p[i] >= 0) gsz(p[i], i, n - s - 1);
      hsz.back()[i] = 0;
    }
  }
  while (q--) {
    int_fast64_t t, v, d, di;
    cin >> t >> v, --v;
    if (t == 1) {
      cin >> d;
      if (g[v].size() > blo) {
        rec[v] += d;
        if (rec[v] >= mo) rec[v] -= mo;
      } else {
        di = d * inv % mo;
        for (register int i = (0); i < (g[v].size()); ++i) {
          int& u = g[v][i];
          int x = di * (n - sz[u]) % mo;
          if (u == p[v]) continue;
          add(lb[u] + 1, x);
          add(rb[u] + 1, mo - x);
        }
        int x = di * sz[v] % mo;
        one += x;
        add(lb[v] + 1, d < x ? mo + d - x : d - x);
        add(rb[v] + 1, x);
        add(lb[v] + 2, mo - d);
      }
    } else {
      int_fast64_t res = 0;
      for (register int i = (0); i < (vt.size()); ++i) {
        res += rec[vt[i]] * (n - hsz[i][v]) % mo;
      }
      cout << (res % mo * inv + sum(lb[v] + 1)) % mo << '\n';
    }
  }
}