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 mod = 998244353;
int add(int a, int b) { return a + b - mod * (a + b >= mod); }
int mul(int a, int b) { return int(1ll * a * b % mod); }
int power(int a, int x) {
  if (x == 0) {
    return 1;
  }
  return mul(power(mul(a, a), x >> 1), (x & 1 ? a : 1));
}
int divide(int a, int b) { return mul(a, power(b, mod - 2)); }
int sub(int a, int b) { return a - b + mod * (a < b); }
int _time = 0, n;
vector<int> tin, tout, f, sz, p;
vector<vector<int>> g;
void update(int k, int x) {
  for (k; k < 2 * n; k |= (k + 1)) {
    f[k] = add(f[k], x);
  }
}
int get(int k) {
  int res = 0;
  for (k; k >= 0; k = (k & (k + 1)) - 1) {
    res = add(res, f[k]);
  }
  return res;
}
void dfs(int v) {
  tin[v] = _time++;
  for (int to : g[v]) {
    if (to == p[v]) {
      continue;
    }
    p[to] = v;
    dfs(to);
    sz[v] += sz[to];
  }
  tout[v] = _time++;
}
const int T = 100;
vector<vector<int>> heavy;
int inv;
void recalc(int v, int pr) {
  sz[v] = 1;
  for (int to : g[v]) {
    if (to == pr) {
      continue;
    }
    recalc(to, v);
    sz[v] += sz[to];
  }
}
void calc(int v, int pr, int id, int subtree = -1) {
  if (pr == -1) {
    heavy[id][v] = 1;
    for (int to : g[v]) {
      calc(to, v, id, sz[to]);
    }
  } else {
    heavy[id][v] = mul(n - subtree, inv);
    for (int to : g[v]) {
      if (to == pr) {
        continue;
      }
      calc(to, v, id, subtree);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int q;
  cin >> n >> q;
  inv = divide(1, n);
  g.resize(n), tin.resize(n), tout.resize(n), f.resize(2 * n), sz.resize(n, 1),
      p.resize(n);
  heavy.resize(T, vector<int>(n));
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  dfs(0);
  vector<int> id(n);
  iota(id.begin(), id.end(), 0);
  sort(id.begin(), id.end(),
       [&](int i, int j) { return g[i].size() > g[j].size(); });
  vector<bool> isHeavy(n);
  vector<int> cnt(n);
  for (int i = 0; i < min(n, T); ++i) {
    isHeavy[id[i]] = true;
    recalc(id[i], -1);
    calc(id[i], -1, i);
  }
  recalc(0, -1);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int v, d;
      cin >> v >> d;
      --v;
      if (isHeavy[v]) {
        cnt[v] = add(cnt[v], d);
        continue;
      }
      for (int to : g[v]) {
        if (to == p[v]) {
          continue;
        }
        int val = mul(d, mul(n - sz[to], inv));
        update(tin[to], val);
        update(tout[to], sub(0, val));
      }
      update(tin[v], d);
      update(tin[v] + 1, sub(0, d));
      if (v == 0) {
        continue;
      }
      int val = mul(d, mul(sz[v], inv));
      update(0, val);
      update(tin[v], sub(0, val));
      update(tout[v], val);
    } else {
      int v;
      cin >> v;
      --v;
      int res = get(tin[v]);
      for (int i = 0; i < min(n, T); ++i) {
        res = add(res, mul(heavy[i][v], cnt[id[i]]));
      }
      cout << res << '\n';
    }
  }
  return 0;
}