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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int N = 150010;
const int p = 998244353;
int n, q;
int head[N], tot;
int siz[N], son[N], fa[N], top[N], lef[N], rig[N], tag[N], cnt;
int sum[N];
struct node {
  int to, next;
} e[N << 1];
void adde(int x, int y) {
  tot++;
  e[tot].to = y;
  e[tot].next = head[x];
  head[x] = tot;
  tot++;
  e[tot].to = x;
  e[tot].next = head[y];
  head[y] = tot;
}
int ksm(int a, int b) {
  int res = 1, base = a;
  while (b) {
    if (b & 1) res = 1ll * res * base % p;
    base = 1ll * base * base % p;
    b >>= 1;
  }
  return res;
}
void dfs1(int u, int f) {
  siz[u] = 1;
  fa[u] = f;
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == f) continue;
    dfs1(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int topf) {
  top[u] = topf;
  lef[u] = ++cnt;
  if (son[u]) dfs2(son[u], topf);
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if ((v == fa[u]) || (v == son[u])) continue;
    dfs2(v, v);
  }
  rig[u] = cnt;
}
int lowbit(int x) { return x & (-x); }
void Add(int x, int d) {
  for (int i = x; i <= n; i += lowbit(i)) sum[i] = (sum[i] + d) % p;
}
int Sum(int x) {
  int res = 0;
  for (int i = x; i >= 1; i -= lowbit(i)) res = (res + sum[i]) % p;
  return res;
}
void modify(int l, int r, int d) {
  Add(l, d);
  Add(r + 1, (p - d));
}
void modify_tree(int u, int d) {
  tag[u] = (tag[u] + d) % p;
  if (son[u]) modify(lef[son[u]], rig[son[u]], 1ll * (n - siz[son[u]]) * d % p);
  if (u != 1) {
    modify(1, lef[u] - 1, 1ll * siz[u] * d % p);
    if (rig[u] != n) modify(rig[u] + 1, n, 1ll * siz[u] * d % p);
  }
}
int query_tree(int u) {
  int res = 1ll * (1ll * n * tag[u] % p + Sum(lef[u])) % p;
  while (u) {
    res = 1ll * (res + 1ll * tag[fa[top[u]]] * (n - siz[top[u]]) % p) % p;
    u = fa[top[u]];
  }
  return res;
}
int main() {
  n = read();
  q = read();
  int inv = ksm(n, p - 2);
  int x, y;
  for (int i = 1; i < n; i++) {
    x = read();
    y = read();
    adde(x, y);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  int op;
  for (int i = 1; i <= q; i++) {
    op = read();
    if (op == 1) {
      x = read();
      y = read();
      modify_tree(x, y);
    } else {
      x = read();
      printf("%d\n", 1ll * query_tree(x) * inv % p);
    }
  }
  return 0;
}