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;
char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
int getnum() {
  int x = 0, f = 1;
  char ch = gc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = gc();
  }
  return x * f;
}
const int N = 150010, mo = 998244353;
int n, Q, B, invn;
int cnt, tim, top;
int head[N];
vector<int> son[N];
int par[N][21];
int st[N];
int C[N];
int W[N];
int dfn[N];
int low[N];
int siz[N];
int depth[N];
void addedge(int u, int v) { son[u].push_back(v); }
void get_read() {
  n = getnum();
  Q = getnum();
  for (int i = 1; i <= n - 1; ++i) {
    int u = getnum(), v = getnum();
    addedge(u, v);
    addedge(v, u);
  }
}
void dfs(int u, int fa) {
  dfn[u] = ++tim;
  depth[u] = depth[fa] + 1;
  siz[u] = 1;
  par[u][0] = fa;
  for (int i = 1; i <= 20; ++i) par[u][i] = par[par[u][i - 1]][i - 1];
  for (int i = 0, L = son[u].size(); i < L; ++i) {
    int v = son[u][i];
    if (v == fa) continue;
    dfs(v, u);
    siz[u] += siz[v];
  }
  low[u] = tim;
}
void add(int &a, int b) {
  a += b;
  if (a >= mo) a -= mo;
}
int ksm(int a, int b) {
  int s = 1;
  while (b) {
    if (b & 1) s = 1LL * s * a % mo;
    a = 1LL * a * a % mo;
    b >>= 1;
  }
  return s;
}
int lowbit(int x) { return x & (-x); }
void modify(int L, int R, int x) {
  for (int i = L; i <= n; i += lowbit(i)) add(C[i], x);
  for (int i = R + 1; i <= n; i += lowbit(i)) add(C[i], mo - x);
}
int query(int p) {
  int ret = 0;
  for (int i = p; i; i -= lowbit(i)) add(ret, C[i]);
  return ret;
}
void calc(int u, int D) {
  for (int i = 0, L = son[u].size(); i < L; ++i) {
    int v = son[u][i];
    if (v == par[u][0]) continue;
    modify(dfn[v], low[v], 1LL * D * (n - siz[v]) % mo * invn % mo);
  }
  modify(1, n, 1LL * D * siz[u] % mo * invn % mo);
  modify(dfn[u], low[u], 1LL * (mo - D) * siz[u] % mo * invn % mo);
}
int get(int u, int D) {
  if (D < 0) return -1;
  for (int i = 20; i >= 0; --i)
    if (D & (1 << i)) u = par[u][i];
  return u;
}
int ask(int u) {
  int ans = (W[u] + query(dfn[u])) % mo;
  for (int i = 1; i <= top; ++i) {
    if (st[i] == u) continue;
    int v = st[i], p = get(u, depth[u] - depth[v] - 1);
    if (p == -1 || par[p][0] != v)
      add(ans, 1LL * W[v] * siz[v] % mo * invn % mo);
    else
      add(ans, 1LL * W[v] * (n - siz[p]) % mo * invn % mo);
  }
  return ans;
}
void get_work() {
  dfs(1, 0);
  B = sqrt(n) + 1;
  invn = ksm(n, mo - 2);
  for (int i = 1; i <= n; ++i)
    if (son[i].size() > B) st[++top] = i;
  while (Q--) {
    int type = getnum();
    if (type == 1) {
      int u = getnum(), D = getnum();
      add(W[u], D);
      if (son[u].size() <= B) calc(u, D);
    } else {
      int u = getnum();
      printf("%d\n", ask(u));
    }
  }
}
int main() {
  get_read();
  get_work();
  return 0;
}