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 n, m;
int sz[maxn + 5], son[maxn + 5], l[maxn + 5], r[maxn + 5], top[maxn + 5],
    fa[maxn + 5];
long long tag[maxn + 5];
int dfstime;
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;
}
long long mul(long long x, long long y) { return x * y % mod; }
void add(long long &x, long long y) { x = (x + y) % mod; }
void dfs(int k, int last) {
  fa[k] = last;
  l[k] = ++dfstime;
  sz[k] = 1;
  son[k] = 0;
  for (auto u : g[k]) {
    if (u == last) continue;
    dfs(u, k);
    sz[k] += sz[u];
    if (sz[u] > sz[son[k]]) son[k] = u;
  }
  r[k] = dfstime;
}
void dfs1(int k, int TOP) {
  top[k] = TOP;
  if (son[k]) dfs1(son[k], TOP);
  for (auto u : g[k]) {
    if (u == fa[k] || u == son[k]) continue;
    dfs1(u, u);
  }
}
namespace BIT {
long long c[maxn + 5];
int lowbit(int x) { return x & (-x); }
void add(int k, long long x) {
  for (; k <= n; k += lowbit(k)) c[k] = (c[k] + x) % mod;
}
long long query(int k) {
  long long ans = 0;
  for (; k; k -= lowbit(k)) ans = (ans + c[k]) % mod;
  return ans;
}
}  // namespace BIT
void add(int l, int r, long long x) {
  if (l > r) return;
  BIT::add(l, x);
  BIT::add(r + 1, -x);
}
long long query(int k) { return BIT::query(k); }
int main() {
  cin >> n >> m;
  long long inv = Pow(n, mod - 2);
  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, 0);
  dfs1(1, 1);
  while (m--) {
    int op, v, d;
    scanf("%d%d", &op, &v);
    if (op == 1) {
      scanf("%d", &d);
      add(1, l[v] - 1, mul(d, sz[v]));
      add(r[v] + 1, n, mul(d, sz[v]));
      add(l[v], l[v], mul(d, n));
      if (son[v]) add(l[son[v]], r[son[v]], mul(d, n - sz[son[v]]));
      add(tag[v], d);
    } else {
      long long ans = query(l[v]);
      v = top[v];
      while (v != 1) {
        add(ans, mul(tag[fa[v]], n - sz[v]));
        v = top[fa[v]];
      }
      ans = ans * inv % mod;
      if (ans < 0) ans += mod;
      printf("%lld\n", ans);
    }
  }
  return 0;
}