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;
constexpr int maxn = 2e5 + 10;
constexpr int threshold = 350;
constexpr int mod = 998244353;
vector<int> G[maxn], son[maxn];
int st[maxn], ed[maxn];
int weight[maxn], bit[maxn];
int fa[maxn], sz[maxn];
int idx;
int qpow(long long a, int n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
void dfs(int u) {
  st[u] = ++idx;
  sz[u] = 1;
  for (auto v : G[u]) {
    if (v == fa[u]) {
      son[u].push_back(son[u].size() ? son[u].back() : 0);
      continue;
    }
    fa[v] = u;
    dfs(v);
    son[u].push_back(st[v]);
    sz[u] += sz[v];
  }
  ed[u] = idx;
}
void upd(int x, int d) {
  if (!x) return;
  for (int i = x; i < maxn; i += i & -i) {
    bit[i] = (bit[i] + d) % mod;
  }
}
int qry(int x) {
  int ret = 0;
  for (int i = x; i; i -= i & -i) {
    ret = (ret + bit[i]) % mod;
  }
  return ret;
}
void rupd(int l, int r, int val) {
  upd(l, val);
  upd(r + 1, mod - val);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1);
  vector<int> heavy;
  for (int i = 1; i <= n; i++) {
    if (G[i].size() >= threshold) heavy.push_back(i);
  }
  int inv_n = qpow(n, mod - 2);
  for (int i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int u, d;
      scanf("%d%d", &u, &d);
      weight[u] = (weight[u] + d) % mod;
      if (G[u].size() < threshold) {
        for (auto v : G[u]) {
          if (v == fa[u]) continue;
          rupd(st[v], ed[v], 1LL * (n - sz[v]) * d % mod * inv_n % mod);
        }
        if (u == 1) continue;
        rupd(1, st[u] - 1, 1LL * sz[u] * d % mod * inv_n % mod);
        rupd(ed[u] + 1, n, 1LL * sz[u] * d % mod * inv_n % mod);
      }
    } else {
      int v;
      scanf("%d", &v);
      int ans = (weight[v] + qry(st[v])) % mod;
      for (auto u : heavy) {
        if (u == v || !weight[u]) continue;
        if (st[v] < st[u] || st[v] > ed[u]) {
          ans = (ans + 1LL * sz[u] * weight[u] % mod * inv_n % mod) % mod;
        } else {
          int id = upper_bound(son[u].begin(), son[u].end(), st[v]) -
                   son[u].begin() - 1;
          int w = G[u][id];
          if (G[u][id] == fa[u]) w = G[u][id - 1];
          ans = (ans + 1LL * (n - sz[w]) * weight[u] % mod * inv_n % mod) % mod;
        }
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}