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 = 150003;
const int mod = 998244353;
int n, q;
int cnt, head[N];
struct Edge {
  int nxt, to;
} e[N << 1];
int dep[N], f[N], siz[N], son[N];
int idx, dfn[N], top[N];
int inv, tag[N], a[N];
int fpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return res;
}
int lowbit(int x) { return x & (-x); }
void modify(int l, int r, int k) {
  for (int i = l; i <= n; i += lowbit(i)) a[i] = (a[i] + k) % mod;
  for (int i = r + 1; i <= n; i += lowbit(i)) a[i] = (a[i] - k + mod) % mod;
}
int query(int i) {
  int res = 0;
  for (; i; i -= lowbit(i)) res = (res + a[i]) % mod;
  return res;
}
void add(int u, int v) {
  e[++cnt] = (Edge){head[u], v}, head[u] = cnt;
  e[++cnt] = (Edge){head[v], u}, head[v] = cnt;
}
void dfs1(int x, int fa) {
  dep[x] = dep[fa] + 1, f[x] = fa, siz[x] = 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to == fa) continue;
    dfs1(to, x);
    siz[x] += siz[to];
    if (siz[to] > siz[son[x]]) son[x] = to;
  }
}
void dfs2(int x, int topf) {
  dfn[x] = ++idx, top[x] = topf;
  if (son[x]) dfs2(son[x], topf);
  for (int i = head[x]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to != f[x] && to != son[x]) dfs2(to, to);
  }
}
void change(int x, int d) {
  tag[x] = (tag[x] + d) % mod;
  if (son[x])
    modify(dfn[son[x]], dfn[son[x]] + siz[son[x]] - 1,
           1ll * (n - siz[son[x]]) * d % mod);
  if (dfn[x] != 1) modify(1, dfn[x] - 1, 1ll * siz[x] * d % mod);
  if (dfn[x] + siz[x] <= n) modify(dfn[x] + siz[x], n, 1ll * siz[x] * d % mod);
}
int qsum(int x) {
  int ans = (tag[x] + 1ll * query(dfn[x]) * inv % mod) % mod;
  while (x) {
    ans = (ans + 1ll * tag[f[top[x]]] * inv % mod * (n - siz[top[x]]) % mod) %
          mod;
    x = f[top[x]];
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  inv = fpow(n, mod - 2);
  while (q--) {
    int o, x, d;
    scanf("%d%d", &o, &x);
    if (o == 1) {
      scanf("%d", &d);
      change(x, d);
    } else
      printf("%d\n", qsum(x));
  }
  return 0;
}