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 = 1.5e5 + 10;
const int mod = 998244353;
const int heavy = 300;
inline int read(void) {
  int ans = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0', ch = getchar();
  return ans;
}
int n, q;
vector<int> g[N], pos[N];
int fa[N], sze[N], dfn[N * 2], m;
int w[2 * N], tree[N * 2];
vector<int> heavy_point;
int ksm(int x, int y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = 1ll * ans * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return ans;
}
void dfs(int u) {
  dfn[++m] = u;
  pos[u].push_back(m);
  sze[u] = 1;
  for (auto v : g[u]) {
    if (v == fa[u]) continue;
    fa[v] = u;
    dfs(v);
    sze[u] += sze[v];
    dfn[++m] = u;
    pos[u].push_back(m);
  }
}
void update(int l, int r, int d) {
  for (int i = l; i <= m; i += i & -i) tree[i] = (tree[i] + d) % mod;
  for (int i = r + 1; i <= m; i += i & -i) tree[i] = (tree[i] - d + mod) % mod;
}
int query(int p) {
  int ans = 0;
  for (int i = p; i >= 1; i -= i & -i) ans = (ans + tree[i]) % mod;
  return ans;
}
int main(void) {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int inv_n = ksm(n, mod - 2);
  dfs(1);
  for (int i = 1; i <= n; ++i) {
    if (g[i].size() >= heavy) heavy_point.push_back(i);
  }
  while (q--) {
    int op, v, d;
    cin >> op >> v;
    if (op == 1) {
      cin >> d;
      w[v] = (w[v] + d) % mod;
      if (g[v].size() >= heavy) continue;
      for (auto u : g[v]) {
        if (u == fa[v]) continue;
        update(pos[u][0], pos[u].back(),
               1ll * d * (n - sze[u]) % mod * inv_n % mod);
      }
      if (v == 1) continue;
      update(1, pos[v][0] - 1, 1ll * d * sze[v] % mod * inv_n % mod);
      update(pos[v].back() + 1, m, 1ll * d * sze[v] % mod * inv_n % mod);
    } else {
      int ans = (w[v] + query(pos[v][0])) % mod;
      for (auto u : heavy_point) {
        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 * w[u] * sze[u] % mod * inv_n % mod) % mod;
        } else {
          int k = dfn[*(--it) + 1];
          ans = (ans + 1ll * w[u] * (n - sze[k]) % mod * inv_n % mod) % mod;
        }
      }
      cout << ans << '\n';
    }
  }
  return 0;
}