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 = 4e5 + 5;
int n, m, mod = 998244353, r, szz;
int tour[N], sz[N], w[N], p[N];
vector<int> heavy;
vector<int> g[N], pos[N];
int f[N];
int add(int a, int b) { return (a + b) % mod; }
int mul(int a, int b) { return (1LL * a * b) % mod; }
void DFS(int i, int j) {
  tour[++r] = i;
  pos[i].push_back(r);
  sz[i] = 1;
  for (int s = 0; s < g[i].size(); ++s) {
    int u = g[i][s];
    if (u != j) {
      DFS(u, i);
      sz[i] += sz[u];
      p[u] = i;
      tour[++r] = i;
      pos[i].push_back(r);
    }
  }
}
int up(int l, int r, int val) {
  for (int i = l; i <= szz; i += i & -i) f[i] = add(f[i], val);
  r++;
  for (int i = r; i <= szz; i += i & -i) f[i] = (f[i] + mod - val) % mod;
}
int get(int i) {
  int ret = 0;
  for (; i; i -= i & -i) ret = add(ret, f[i]);
  return ret;
}
int powmod(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
void sp(int &num) {
  num = 0;
  register int c;
  for (c = getchar(); c < 48 || c > 57; c = getchar())
    ;
  for (; c >= 48 && c <= 57; c = getchar()) num = num * 10 + c - 48;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  sp(n);
  sp(m);
  for (int i = 1; i <= n - 1; ++i) {
    int x, y;
    sp(x);
    sp(y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int inv = powmod(n, mod - 2);
  DFS(1, -1);
  int k = 300;
  for (int i = 1; i <= n; ++i)
    if (g[i].size() >= k) heavy.push_back(i);
  szz = 2 * n - 1;
  for (int i = 1; i <= m; ++i) {
    int type, u, val;
    sp(type);
    if (type == 1) {
      sp(u);
      sp(val);
      w[u] = add(w[u], val);
      if (g[u].size() < k) {
        for (int s = 0; s < g[u].size(); ++s) {
          int v = g[u][s];
          if (v != p[u])
            up(pos[v][0], pos[v].back(), mul(mul(n - sz[v], val), inv));
        }
        if (u == 1) continue;
        up(1, pos[u][0] - 1, mul(mul(sz[u], val), inv));
        up(pos[u].back() + 1, szz, mul(mul(sz[u], val), inv));
      }
    } else {
      sp(u);
      int ans = 0;
      ans = add(w[u], get(pos[u][0]));
      for (int s = 0; s < heavy.size(); ++s) {
        int v = heavy[s];
        if (u == v) continue;
        auto it = lower_bound(pos[v].begin(), pos[v].end(), pos[u][0]);
        if (it == pos[v].begin() || it == pos[v].end())
          ans = add(ans, mul(mul(sz[v], w[v]), inv));
        else {
          int p = tour[*(--it) + 1];
          ans = add(ans, mul(mul(n - sz[p], w[v]), inv));
        }
      }
      cout << ans << "\n";
    }
  }
}