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 maxn = 150005;
const int mod = 998244353;
const int heavy = 300;
int n, m, q;
vector<int> adj[maxn];
int dad[maxn];
int sz[maxn];
vector<int> pos[maxn];
int tour[2 * maxn];
vector<int> heavy_vector;
int weight[2 * maxn];
int pw(int x, int y) {
  int r = 1;
  while (y) {
    if (y & 1) r = 1LL * r * x % mod;
    x = 1LL * x * x % mod;
    y >>= 1;
  }
  return r;
}
void dfs(int u) {
  tour[++m] = u;
  pos[u].push_back(m);
  sz[u] = 1;
  for (auto v : adj[u]) {
    if (v == dad[u]) continue;
    dad[v] = u;
    dfs(v);
    sz[u] += sz[v];
    tour[++m] = u;
    pos[u].push_back(m);
  }
}
int fwt[2 * maxn];
void upd(int l, int r, int d) {
  for (int p = l; p <= m; p += p & -p) fwt[p] = (fwt[p] + d) % mod;
  ++r;
  for (int p = r; p <= m; p += p & -p) fwt[p] = (fwt[p] + mod - d) % mod;
}
int get(int p) {
  int ret = 0;
  for (; p; p -= p & -p) ret = (ret + fwt[p]) % mod;
  return ret;
}
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> q;
  int inv_n = pw(n, mod - 2);
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  for (int i = 1; i <= n; ++i)
    if (adj[i].size() >= heavy) heavy_vector.push_back(i);
  while (q--) {
    int type, v;
    cin >> type >> v;
    if (type == 1) {
      int d;
      cin >> d;
      weight[v] = (weight[v] + d) % mod;
      if (adj[v].size() < heavy) {
        for (auto u : adj[v])
          if (u != dad[v])
            upd(pos[u][0], pos[u].back(),
                1LL * d * (n - sz[u]) % mod * inv_n % mod);
        if (v == 1) continue;
        upd(1, pos[v][0] - 1, 1LL * d * sz[v] % mod * inv_n % mod);
        upd(pos[v].back() + 1, m, 1LL * d * sz[v] % mod * inv_n % mod);
      }
    } else {
      int ans = (weight[v] + get(pos[v][0])) % mod;
      for (auto u : heavy_vector) {
        if (u == v) continue;
        auto it = lower_bound(pos[u].begin(), pos[u].end(), pos[v][0]);
        if (it == pos[u].begin() || it == pos[u].end())
          ans = (ans + 1LL * weight[u] * sz[u] % mod * inv_n) % mod;
        else {
          int p = tour[*(--it) + 1];
          ans = (ans + 1LL * weight[u] * (n - sz[p]) % mod * inv_n) % mod;
        }
      }
      cout << ans << '\n';
    }
  }
  return 0;
}