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 = 150005;
int n, q;
int inv[N];
struct Edge {
  int num;
  int next;
} edge[2 * N];
long long c[N];
int tot, last[N];
int cnt[N], son[N];
int top[N], fa[N];
int L[N], R[N], now;
int w[N];
const int mod = 998244353;
void Add(int i, int j) {
  tot++;
  edge[tot].num = j;
  edge[tot].next = last[i];
  last[i] = tot;
}
void dfs(int u, int father) {
  fa[u] = father;
  int maxn = 0;
  for (int k = last[u]; k != 0; k = edge[k].next) {
    int v = edge[k].num;
    if (v == father) continue;
    dfs(v, u);
    cnt[u] += cnt[v];
    if (cnt[v] > maxn) {
      maxn = cnt[v];
      son[u] = v;
    }
  }
  cnt[u]++;
}
void dfs1(int u, int head, int father) {
  top[u] = head;
  L[u] = ++now;
  if (son[u]) dfs1(son[u], head, u);
  for (int k = last[u]; k != 0; k = edge[k].next) {
    int v = edge[k].num;
    if (v == son[u]) continue;
    if (v == father) continue;
    dfs1(v, v, u);
  }
  R[u] = now;
}
inline int lowbit(int x) { return x & (-x); }
void add(int i, int p) {
  for (int j = i; j <= n; j += lowbit(j)) {
    c[j] = c[j] + p;
    if (c[j] >= mod) c[j] -= mod;
    if (c[j] < 0) c[j] += mod;
  }
}
int Find(int i) {
  int ans = 0;
  for (int j = i; j > 0; j -= lowbit(j)) {
    ans = ans + c[j];
    if (ans >= mod) ans -= mod;
    if (ans < 0) ans += mod;
  }
  return ans;
}
void Add(int l, int r, int p) {
  add(l, p);
  add(r + 1, -p);
}
int main() {
  scanf("%d %d", &n, &q);
  inv[0] = 1;
  inv[1] = 1;
  for (int i = 2; i <= n; i++)
    inv[i] = (1ll * (mod - mod / i) * inv[mod % i]) % mod;
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    Add(u, v);
    Add(v, u);
  }
  dfs(1, 0);
  dfs1(1, 1, 0);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int v, d;
      scanf("%d %d", &v, &d);
      Add(1, L[v] - 1, 1ll * cnt[v] * d % mod * inv[n] % mod);
      Add(R[v] + 1, n, 1ll * cnt[v] * d % mod * inv[n] % mod);
      if (son[v])
        Add(L[son[v]], R[son[v]],
            1ll * (n - cnt[son[v]]) * d % mod * inv[n] % mod);
      w[v] = (w[v] + d) % mod;
    } else {
      int v;
      long long ans = 0;
      scanf("%d", &v);
      ans = Find(L[v]);
      for (int x = top[v]; x != 0; x = top[fa[x]])
        ans = (ans + 1ll * (n - cnt[x]) * w[fa[x]] % mod * inv[n] % mod) % mod;
      ans = (ans + w[v]) % mod;
      printf("%d\n", ans);
    }
  }
}