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;
int N, Q;
const long long MOD = 998244353;
const int B = 200;
vector<int> graph[150005];
long long val[150005];
vector<int> hvy;
bool ishvy[150005];
int sz[150005];
int lft[150005], rht[150005], t;
long long bit[150005];
int par[150005];
vector<pair<int, int>> lst[150005];
long long mult(long long a, long long b) { return a * b % MOD; }
long long add(long long a, long long b) { return (a + b) % MOD; }
long long qikpow(long long b, long long e) {
  if (!e) {
    return 1;
  }
  long long ret = qikpow(b, e / 2);
  ret = mult(ret, ret);
  if (e & 1) {
    ret = mult(ret, b);
  }
  return ret;
}
void dfs(int n) {
  lft[n] = ++t;
  sz[n] = 1;
  for (int e : graph[n]) {
    if (e != par[n]) {
      par[e] = n;
      dfs(e);
      sz[n] += sz[e];
    }
  }
  rht[n] = t;
}
void upd(int n, long long v) {
  if (v < 0) {
    v += MOD;
  }
  while (n <= N) {
    bit[n] = add(bit[n], v);
    n += n & -n;
  }
}
long long query(int n) {
  long long ret = 0;
  while (n) {
    ret = add(ret, bit[n]);
    n -= n & -n;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> Q;
  for (int i = 1; i < N; i++) {
    int a, b;
    cin >> a >> b;
    graph[a].emplace_back(b);
    graph[b].emplace_back(a);
  }
  for (int i = 1; i <= N; i++) {
    if (graph[i].size() > B) {
      hvy.emplace_back(i);
      ishvy[i] = 1;
    }
  }
  dfs(1);
  const long long invN = qikpow(N, MOD - 2);
  for (int n : hvy) {
    for (int e : graph[n]) {
      if (e != par[n]) {
        lst[n].emplace_back(lft[e], sz[e]);
      }
    }
  }
  while (Q--) {
    int c, n, d;
    cin >> c >> n;
    if (c == 1) {
      cin >> d;
      val[n] = add(val[n], d);
      if (!ishvy[n]) {
        upd(1, d - mult(d, mult(invN, N - sz[n])));
        upd(lft[n], mult(d, mult(invN, N - sz[n])));
        upd(rht[n] + 1, -mult(d, mult(invN, N - sz[n])));
        for (int e : graph[n]) {
          if (e != par[n]) {
            upd(lft[e], -mult(d, mult(invN, sz[e])));
            upd(rht[e] + 1, mult(d, mult(invN, sz[e])));
          }
        }
      }
    } else {
      long long out = query(lft[n]);
      for (int k : hvy) {
        if (k == n) {
          out = add(out, val[k]);
        } else if (lft[k] < lft[n] && rht[k] >= rht[n]) {
          auto it = --upper_bound(lst[k].begin(), lst[k].end(),
                                  make_pair(lft[n], INT_MAX));
          out = add(out, mult(val[k], mult(invN, N - it->second)));
        } else {
          out = add(out, mult(val[k], mult(invN, sz[k])));
        }
      }
      cout << out << "\n";
    }
  }
}