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 = 1000005;
const int mod = 998244353;
void inc(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int qpow(int a, int b) {
  int tmp = 1;
  while (b) {
    if (b & 1) tmp = 1ll * tmp * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return tmp;
}
int n, ni, q, hd[N], cnt, siz[N], fa[N], dep[N], son[N], top[N], dfn[N], dfc,
    w[N];
struct Edge {
  int to, nxt;
} e[N << 1];
void adde(int u, int v) {
  e[++cnt].to = v;
  e[cnt].nxt = hd[u];
  hd[u] = cnt;
  e[++cnt].to = u;
  e[cnt].nxt = hd[v];
  hd[v] = cnt;
}
void dfs1(int u) {
  siz[u] = 1;
  for (int i = hd[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa[u]) continue;
    fa[v] = u;
    dep[v] = dep[u] + 1;
    dfs1(v);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int topf) {
  top[u] = topf;
  dfn[u] = ++dfc;
  if (!son[u]) return;
  dfs2(son[u], topf);
  for (int i = hd[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa[u] || v == son[u]) continue;
    dfs2(v, v);
  }
}
namespace BIT {
int c[N];
void modify(int x, int v) {
  for (int i = x; i <= n; i += (i & -i)) {
    inc(c[i], v);
  }
}
int query(int x) {
  long long sum = 0;
  for (int i = x; i >= 1; i ^= (i & -i)) {
    sum += c[i];
  }
  return sum % mod;
}
}  // namespace BIT
void add(int l, int r, int v) {
  BIT::modify(l, v);
  BIT::modify(r + 1, mod - v);
}
void modifysubtree(int x, int d) { add(dfn[x], dfn[x] + siz[x] - 1, d); }
void modify(int x, int d) {
  inc(w[x], d);
  if (son[x]) modifysubtree(son[x], 1ll * (n - siz[son[x]]) * d % mod);
  if (!fa[x]) return;
  modifysubtree(1, 1ll * siz[x] * d % mod);
  modifysubtree(x, 1ll * siz[x] * (mod - d) % mod);
}
void query(int x) {
  int sum = BIT::query(dfn[x]);
  sum = (sum + 1ll * n * w[x]) % mod;
  while (x) {
    sum = (sum + 1ll * w[fa[top[x]]] * (n - siz[top[x]])) % mod;
    x = fa[top[x]];
  }
  printf("%d\n", 1ll * sum * ni % mod);
}
int main() {
  scanf("%d%d", &n, &q);
  ni = qpow(n, mod - 2);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adde(u, v);
  }
  dfs1(1);
  dfs2(1, 1);
  for (int i = 1; i <= q; ++i) {
    int op, x, d;
    scanf("%d%d", &op, &x);
    if (op == 1) {
      scanf("%d", &d);
      modify(x, d);
    } else
      query(x);
  }
  return 0;
}