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 = 200010;
int n, u, v, d, q;
int pre[N];
bool done[N];
int tmp1, tmp2, tmp3, tmp4, p;
int fres = 1;
vector<int> g[N];
int sz2[N];
int pp[N];
void solve(int node) {
  sz2[node] = 1;
  for (int i = 0; i < (int)g[node].size(); i++) {
    if (g[node][i] == pp[node]) continue;
    pp[g[node][i]] = node;
    solve(g[node][i]);
    sz2[node] += sz2[g[node][i]];
  }
}
inline int fastpower(int num, int po) {
  fres = 1;
  while (po > 0) {
    if (po & 1) fres = (long long)fres * num % 998244353;
    po >>= 1;
    num = (long long)num * num % 998244353;
  }
  return fres;
}
struct S {
  vector<int> v;
  vector<int> a;
  int all = 0, all2 = 0, all3 = 0;
  int cur;
  int sz;
  vector<pair<int, pair<int, int> > > v2;
  void update(int node, int d) {
    if (node == cur) {
      all3 += d;
      if (all3 >= 998244353) all3 -= 998244353;
      all2 += (long long)d * n % 998244353;
      if (all2 >= 998244353) all2 -= 998244353;
      return;
    }
    tmp1 = lower_bound(v2.begin(), v2.end(), make_pair(node, make_pair(0, 0))) -
           v2.begin();
    assert(v2[tmp1].first == node);
    tmp2 = v2[tmp1].second.first;
    all += (long long)d * v2[tmp1].second.second % 998244353;
    if (all >= 998244353) all -= 998244353;
    a[tmp2] += (long long)d * v2[tmp1].second.second % 998244353;
    if (a[tmp2] >= 998244353) a[tmp2] -= 998244353;
  }
  int get(int node) {
    if (node == cur) {
      return (all2 + all) % 998244353;
    }
    tmp1 = lower_bound(v2.begin(), v2.end(), make_pair(node, make_pair(0, 0))) -
           v2.begin();
    assert(node == v2[tmp1].first);
    tmp2 = v2[tmp1].second.first;
    tmp3 = all - a[tmp2];
    if (tmp3 < 0) tmp3 += 998244353;
    tmp3 += (long long)all3 * (n - v[tmp2]) % 998244353;
    if (tmp3 >= 998244353) tmp3 -= 998244353;
    return tmp3;
  }
};
S arr[N];
int sz[N];
void init(int node, int prnt) {
  sz[node] = 1;
  for (int i = 0; i < (int)g[node].size(); i++) {
    if (g[node][i] != prnt && !done[g[node][i]]) {
      init(g[node][i], node);
      sz[node] += sz[g[node][i]];
    }
  }
}
int all = 0;
int get_centroid(int node, int prnt) {
  for (int i = 0; i < (int)g[node].size(); i++) {
    if (g[node][i] == prnt || done[g[node][i]]) continue;
    if (sz[g[node][i]] > all / 2) return get_centroid(g[node][i], node);
  }
  return node;
}
int Cur;
int cnt = 0;
void DFS(int node, int prnt, int val) {
  for (int i = 0; i < (int)g[node].size(); i++) {
    if (done[g[node][i]] || g[node][i] == prnt) continue;
    if (pp[node] == g[node][i])
      DFS(g[node][i], node, sz2[node]);
    else
      DFS(g[node][i], node, n - sz2[g[node][i]]);
  }
  arr[Cur].v2.push_back(make_pair(node, make_pair(cnt, n - val)));
}
void calc(int node, int last) {
  init(node, -1);
  all = sz[node];
  node = get_centroid(node, -1);
  pre[node] = last;
  cnt = 0;
  Cur = node;
  arr[node].sz = all;
  for (int i = 0; i < (int)g[node].size(); i++) {
    if (done[g[node][i]]) continue;
    if (pp[g[node][i]] == node) {
      arr[node].v.push_back(sz2[g[node][i]]);
      DFS(g[node][i], node, n - sz2[g[node][i]]);
    } else {
      arr[node].v.push_back(n - sz2[node]);
      DFS(g[node][i], node, sz2[node]);
    }
    cnt++;
  }
  arr[node].a.resize(cnt, 0);
  arr[node].all = 0;
  arr[node].all2 = 0;
  arr[node].all3 = 0;
  arr[node].cur = node;
  done[node] = true;
  for (int i = 0; i < (int)g[node].size(); i++) {
    if (done[g[node][i]]) continue;
    calc(g[node][i], node);
  }
  sort(arr[node].v2.begin(), arr[node].v2.end());
}
int main() {
  scanf("%d%d", &n, &q);
  p = fastpower(n, 998244353 - 2) % 998244353;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  solve(1);
  calc(1, -1);
  int t, ans = 0, start = 0;
  while (q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &u, &d);
      start = u;
      while (u != -1) {
        arr[u].update(start, d);
        u = pre[u];
      }
    } else {
      scanf("%d", &u);
      start = u;
      ans = 0;
      while (u != -1) {
        ans += arr[u].get(start);
        if (ans >= 998244353) ans -= 998244353;
        u = pre[u];
      }
      ans = (long long)ans * p % 998244353;
      printf("%d\n", ans);
    }
  }
  return 0;
}