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>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int MAXN = 150 * 1000 + 2, MOD = 998244353, N = MAXN * 4, L = 21;
int n, q, ft[MAXN], st[MAXN], tim, sub[MAXN], par[MAXN], pt[MAXN], lg[MAXN];
long long seg[N], sum[MAXN], inv;
vector<int> adj[MAXN];
vector<int> ch[MAXN];
bool hv[MAXN];
inline long long add(long long a, long long b) {
  return ((a % MOD + b % MOD) % MOD + MOD) % MOD;
}
inline long long mul(long long a, long long b) {
  return (a % MOD * b % MOD) % MOD;
}
inline long long Pow(long long a, long long b) {
  if (b == 0) return 1ll;
  long long x = Pow(a, b / 2);
  x = mul(x, x);
  if (b % 2) x = mul(x, a);
  return x;
}
inline void dfs(int v, int p) {
  par[v] = p;
  sub[v] = 1;
  st[v] = tim++;
  lg[v] = -1;
  for (int u : adj[v])
    if (u != p) {
      dfs(u, v);
      if (lg[v] == -1 || sub[lg[v]] < sub[u]) lg[v] = u;
      sub[v] += sub[u];
    }
  ft[v] = tim;
  return;
}
inline void _dfs(int v, int p) {
  if (lg[v] != -1) {
    hv[lg[v]] = 1;
    pt[lg[v]] = pt[v];
  }
  for (auto u : adj[v])
    if (u != p) {
      _dfs(u, v);
    }
  return;
}
inline void upd(long long val, int l, int r) {
  l += n;
  r += n;
  while (l < r) {
    if (l & 1) {
      seg[l] = add(seg[l], val);
      l++;
    }
    if (r & 1) {
      seg[r - 1] = add(seg[r - 1], val);
    }
    l >>= 1;
    r >>= 1;
  }
  return;
}
inline long long get(int x) {
  x += n;
  long long ret = 0;
  for (int i = x; i > 0; i >>= 1) {
    ret = add(ret, seg[i]);
  }
  return ret;
}
int32_t main() {
  iota(pt, pt + MAXN, 0);
  scanf("%d %d", &n, &q);
  inv = Pow(n, MOD - 2);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0, -1);
  _dfs(0, -1);
  while (q--) {
    int t, v;
    long long d;
    scanf("%d %d", &t, &v);
    v--;
    if (t == 1) {
      scanf("%I64d", &d);
      sum[v] = add(sum[v], d);
      int u = lg[v];
      if (v != 0) {
        upd(mul(d, sub[v]), 0, tim);
        upd(mul(-d, sub[v]), st[v], ft[v]);
      }
      if (u != -1) upd(mul(d, (n - sub[u])), st[u], ft[u]);
    } else {
      long long ans = add(sum[v], mul(get(st[v]), inv));
      int u = v;
      while (u != -1) {
        u = pt[u];
        if (par[u] == -1) break;
        ans = add(ans, mul(mul(sum[par[u]], (n - sub[u])), inv));
        u = par[u];
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}