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 long long mod = 998244353;
const int maxn = 150000;
vector<int> g[maxn + 5];
int sz[maxn + 5], dep[maxn + 5];
int fa[18][maxn + 5];
int tag[maxn + 5], ans[maxn + 5], add[maxn + 5];
long long inv;
int B;
vector<pair<int, int>> q;
int n, m;
template <class T>
inline void rd(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 - '0' + c, c = getchar();
  x *= f;
}
namespace Tree {
int len[maxn + 5], son[maxn + 5], top[maxn + 5];
int Log[maxn + 5];
vector<int> up[maxn + 5], down[maxn + 5];
void dfs(int k) {
  for (auto u : g[k]) {
    if (u == fa[0][k]) continue;
    dfs(u);
    if (len[u] > len[k]) son[k] = u;
    len[k] = max(len[u], len[k]);
  }
  len[k] += 1;
}
void dfs1(int k, int TOP) {
  top[k] = TOP;
  if (son[k]) dfs1(son[k], TOP);
  for (auto u : g[k]) {
    if (u == fa[0][k] || u == son[k]) continue;
    dfs1(u, u);
  }
}
void init() {
  Log[1] = 0;
  for (int i = 2; i <= n; ++i) Log[i] = Log[i >> 1] + 1;
  dfs(1);
  dfs1(1, 1);
  for (int i = 1; i <= n; ++i)
    if (top[i] == i) {
      up[i].resize(len[i]);
      down[i].resize(len[i]);
      for (int j = 0, u = i; j < len[i]; ++j, u = son[u]) down[i][j] = u;
      for (int j = 0, u = i; j < len[i]; ++j, u = fa[0][u]) up[i][j] = u;
    }
}
int query(int u, int k) {
  if (dep[u] <= k) return 0;
  if (k == 0) return u;
  int tmp = Log[k];
  u = fa[tmp][u];
  k -= 1 << tmp;
  if (dep[u] - dep[top[u]] >= k) return down[top[u]][dep[u] - dep[top[u]] - k];
  k -= dep[u] - dep[top[u]];
  return up[top[u]][k];
}
}  // namespace Tree
long long Pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
void dfs(int k, int last) {
  fa[0][k] = last;
  dep[k] = dep[last] + 1;
  sz[k] = 1;
  for (auto u : g[k])
    if (u != last) dfs(u, k), sz[k] += sz[u];
}
int query(int u, int v) {
  if (u == v) return n;
  if (dep[u] <= dep[v]) return sz[v];
  int d = dep[u] - dep[v] - 1;
  u = Tree::query(u, d);
  if (fa[0][u] == v)
    return n - sz[u];
  else
    return sz[v];
}
int query(int u) {
  int res = ans[u];
  for (auto v : q) res = (res + 1LL * query(u, v.first) * v.second % mod) % mod;
  if (res < 0) res += mod;
  return res;
}
void dfs1(int k, int sum) {
  ans[k] = (ans[k] + sum) % mod;
  for (auto u : g[k]) {
    if (u == fa[0][k]) continue;
    dfs1(u, (sum + add[k] + 1LL * tag[k] * sz[u]) % mod);
  }
}
void rebuild() {
  int TAG = 0;
  for (auto tmp : q) {
    int v = tmp.first;
    TAG = (TAG + 1LL * tmp.second * sz[v]) % mod;
    tag[v] = (tag[v] - tmp.second) % mod;
    add[v] = (add[v] + 1LL * (n - sz[v]) * tmp.second) % mod;
  }
  dfs1(1, 0);
  for (int i = 1; i <= n; ++i)
    ans[i] = (1LL * ans[i] + TAG + add[i]) % mod, tag[i] = add[i] = 0;
  q.clear();
}
int main() {
  rd(n), rd(m);
  inv = Pow(n, mod - 2);
  B = 600;
  for (int i = 1; i < n; ++i) {
    int u, v;
    rd(u), rd(v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  for (int j = 1; j < 18; ++j)
    for (int i = 1; i <= n; ++i) fa[j][i] = fa[j - 1][fa[j - 1][i]];
  Tree::init();
  while (m--) {
    int op, v, d;
    rd(op);
    if (op == 1) {
      rd(v), rd(d);
      q.push_back(make_pair(v, d));
      if (q.size() > B) rebuild();
    } else {
      rd(v);
      int ans = 1LL * query(v) * inv % mod;
      if (ans < 0) ans += mod;
      printf("%d\n", ans);
    }
  }
  return 0;
}