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 = 1e6 + 10;
const int mod = 998244353;
int inc(int x, int y) { return (x + y >= mod) ? (x + y - mod) : (x + y); }
int del(int x, int y) { return (x - y < 0) ? (x - y + mod) : (x - y); }
const int SEN = 2e5 + 10;
struct Segment {
  int sum[SEN << 2], lazy[SEN << 2];
  void addlazy(int pos, int l, int r, int w) {
    sum[pos] = inc(sum[pos], 1ll * w * (r - l + 1) % mod);
    lazy[pos] = inc(lazy[pos], w);
  }
  void pushup(int pos) { sum[pos] = inc(sum[pos << 1], sum[pos << 1 | 1]); }
  void pushdown(int pos, int l, int r) {
    if (lazy[pos] != 0) {
      int mid = (l + r) >> 1;
      addlazy(pos << 1, l, mid, lazy[pos]);
      addlazy(pos << 1 | 1, mid + 1, r, lazy[pos]);
      lazy[pos] = 0;
    }
  }
  int query(int ql, int qr, int pos, int l, int r) {
    if (ql <= l && r <= qr) {
      return sum[pos];
    }
    pushdown(pos, l, r);
    int mid = (l + r) >> 1;
    int ans = 0;
    if (ql <= mid) ans = inc(ans, query(ql, qr, pos << 1, l, mid));
    if (qr > mid) ans = inc(ans, query(ql, qr, pos << 1 | 1, mid + 1, r));
    return ans;
  }
  void update(int ql, int qr, int w, int pos, int l, int r) {
    if (ql > qr) return;
    if (ql <= l && r <= qr) {
      addlazy(pos, l, r, w);
      return;
    }
    pushdown(pos, l, r);
    int mid = (l + r) >> 1;
    if (ql <= mid) update(ql, qr, w, pos << 1, l, mid);
    if (qr > mid) update(ql, qr, w, pos << 1 | 1, mid + 1, r);
    pushup(pos);
  }
} t;
namespace treepo {
int top[N], son[N], dep[N], f[N], siz[N], dfn[N], ed[N];
vector<int> g[N];
int tag[N];
int tot;
void dfs1(int x, int fa) {
  dep[x] = dep[fa] + 1;
  siz[x] = 1;
  f[x] = fa;
  for (int to : g[x]) {
    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 fa, int tp) {
  top[x] = tp;
  dfn[x] = ++tot;
  if (son[x] != 0) {
    dfs2(son[x], x, tp);
  }
  for (int to : g[x]) {
    if (to == fa) continue;
    if (to == fa || to == son[x]) continue;
    dfs2(to, x, to);
  }
  ed[x] = tot;
}
int query(int u) {
  int res = inc(1ll * tot * tag[u] % mod, t.query(dfn[u], dfn[u], 1, 1, tot));
  while (u) {
    res = inc(res, 1ll * tag[f[top[u]]] * (tot - siz[top[u]]) % mod);
    u = f[top[u]];
  }
  return res;
}
void update(int u, int d) {
  tag[u] = inc(tag[u], d);
  if (son[u])
    t.update(dfn[son[u]], ed[son[u]], 1ll * d * (tot - siz[son[u]]) % mod, 1, 1,
             tot);
  t.update(1, dfn[u] - 1, 1ll * d * siz[u] % mod, 1, 1, tot);
  t.update(ed[u] + 1, tot, 1ll * d * siz[u] % mod, 1, 1, tot);
}
void init(int n) {
  dfs1(1, 0);
  dfs2(1, 0, 1);
}
}  // namespace treepo
using namespace treepo;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return x * f;
}
int qpow(int a, int x, int mo) {
  int res = 1;
  while (x) {
    if (x & 1) res = 1ll * res * a % mo;
    a = 1ll * a * a % mo;
    x >>= 1;
  }
  return res;
}
int main() {
  int n = read(), q = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  init(n);
  int inv = qpow(n, mod - 2, mod);
  for (int i = 1; i <= q; i++) {
    int op = read();
    if (op == 1) {
      int u = read(), d = read();
      update(u, d);
    } else {
      int u = read();
      cout << 1ll * query(u) * inv % mod << endl;
    }
  }
}