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 b = 230;
const int mod = 998244353;
long long pow2(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b % 2 == 1) ans = ans * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return ans;
}
vector<int> g[150005];
int par[18][150005];
int depth[150005];
long long sum[150005];
int sz[150005];
int l[150005], r[150005];
int tim = 0;
void dfs(int u, int p, int d) {
  sz[u] = 1;
  par[0][u] = p;
  depth[u] = d;
  l[u] = ++tim;
  for (int i = 0; i < (int)(g[u]).size(); i++) {
    int v = g[u][i];
    if (v != p) {
      dfs(v, u, d + 1);
      sz[u] += sz[v];
    }
  }
  r[u] = tim;
}
long long bit[150005];
void add(int idx, long long val) {
  for (++idx; idx <= 150000; idx += idx & -idx) {
    bit[idx] = (bit[idx] + val) % mod;
    if (bit[idx] >= mod) bit[idx] -= mod;
  }
}
void range_add(int l, int r, long long val) {
  add(l, val);
  add(r + 1, (-val + mod) % mod);
}
long long point_query(int idx) {
  long long ret = 0;
  for (++idx; idx > 0; idx -= idx & -idx) {
    ret = (ret + bit[idx]) % mod;
    if (ret >= mod) ret -= mod;
  }
  return ret;
}
int jump(int u, int d) {
  for (int i = 17; i >= 0; i--) {
    if ((d >> i) & 1) u = par[i][u];
  }
  return u;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].emplace_back(b);
    g[b].emplace_back(a);
  }
  vector<int> heavy;
  for (int i = 1; i <= n; i++) {
    if ((int)(g[i]).size() > b) heavy.emplace_back(i);
  }
  long long inv = pow2(n, mod - 2);
  dfs(1, 0, 0);
  for (int j = 1; j < 18; j++) {
    for (int i = 1; i <= n; i++) {
      if (par[j - 1][i] != 0) par[j][i] = par[j - 1][par[j - 1][i]];
    }
  }
  for (int i = 1; i <= q; i++) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int u, d;
      scanf("%d%d", &u, &d);
      if ((int)(g[u]).size() <= b) {
        range_add(l[u], l[u], d);
        for (int i = 0; i < (int)(g[u]).size(); i++) {
          int v = g[u][i];
          if (sz[u] > sz[v]) {
            long long sub = n - sz[v];
            long long p = sub * 1ll * inv % mod;
            range_add(l[v], r[v], p * 1ll * d % mod);
          } else {
            long long sub = sz[u];
            long long p = sub * 1ll * inv % mod;
            if (1 <= l[u] - 1) range_add(1, l[u] - 1, p * 1ll * d % mod);
            if (r[u] + 1 <= n) range_add(r[u] + 1, n, p * 1ll * d % mod);
          }
        }
      } else {
        sum[u] = (sum[u] + d);
        if (sum[u] >= mod) sum[u] -= mod;
      }
    } else {
      int u;
      scanf("%d", &u);
      long long ans = point_query(l[u]);
      for (int i = 0; i < (int)(heavy).size(); i++) {
        int v = heavy[i];
        if (l[v] <= l[u] && r[u] <= r[v]) {
          if (u == v)
            ans = (ans + sum[v]) % mod;
          else {
            int u2 = jump(u, depth[u] - depth[v] - 1);
            long long sub = n - sz[u2];
            long long p = sub * 1ll * inv % mod;
            ans = (ans + p * 1ll * sum[v] % mod);
            if (ans >= mod) ans -= mod;
          }
        } else {
          long long sub = sz[v];
          long long p = sub * 1ll * inv % mod;
          ans = (ans + p * 1ll * sum[v] % mod);
          if (ans >= mod) ans -= mod;
        }
      }
      printf("%lld\n", ans);
    }
  }
}