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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long fpow(long long b, long long p) {
  if (b == 0) return 0;
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * b % (998244353);
    b = b * b % (998244353);
    p /= 2;
  }
  return ans;
}
struct BIT {
  long long n;
  long long a[150050];
  void init(long long _n) {
    n = _n;
    memset(a, 0, sizeof(a));
  }
  void add(long long i, long long x) {
    while (i <= n) {
      a[i] += x;
      if (a[i] >= (998244353)) a[i] -= (998244353);
      i += (i & (-i));
    }
  }
  long long ask(long long l, long long r) {
    long long ans = 0;
    while (r > 0) {
      ans += a[r];
      r -= (r & (-r));
    }
    l--;
    while (l > 0) {
      ans -= a[l];
      l -= (l & (-l));
    }
    return ans % (998244353);
  }
} bit;
long long N, N1;
long long addall = 0;
vector<long long> adj[150010];
long long sz[150010];
long long p[150010];
long long nxt[150010];
long long id[150010];
long long head[150010];
long long add[150010];
long long info(long long n, long long par) {
  sz[n] = 1;
  p[n] = par;
  pair<long long, long long> mx(0, 0);
  for (auto &i : adj[n])
    if (i != par) {
      mx = max(mx, make_pair(info(i, n), i));
      sz[n] += sz[i];
    }
  nxt[n] = mx.second;
  return sz[n];
}
void hld(long long n, long long h) {
  static long long nowid = 1;
  id[n] = nowid;
  nowid++;
  head[n] = h;
  if (nxt[n] != 0) hld(nxt[n], h);
  for (auto &i : adj[n])
    if (i != p[n] && i != nxt[n]) {
      hld(i, i);
    }
}
long long ask(long long n) {
  long long ans = addall + add[n];
  while (n != 0) {
    ans += bit.ask(id[head[n]], id[n]);
    n = head[n];
    ans = (ans + add[p[n]] * (N - sz[n]) % (998244353) * N1) % (998244353);
    n = p[n];
  }
  return ans;
}
signed main() {
  {
    ;
    ;
    ;
    ios::sync_with_stdio(false);
    cin.tie(0);
    ;
    ;
    ;
  }
  cin >> N;
  long long q;
  cin >> q;
  N1 = fpow(N, (998244353) - 2);
  for (long long i = 1; i <= N - 1; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].emplace_back(b);
    adj[b].emplace_back(a);
  }
  bit.init(N);
  info(1, 0);
  hld(1, 1);
  while (q--) {
    long long op;
    cin >> op;
    if (op == 1) {
      long long v, x;
      cin >> v >> x;
      addall = (addall + x * sz[v] % (998244353) * N1) % (998244353);
      add[v] = (add[v] + x) % (998244353);
      bit.add(id[v], -x * sz[v] % (998244353) * N1 % (998244353));
      if (nxt[v] != 0) {
        bit.add(id[nxt[v]],
                x * (N - sz[nxt[v]]) % (998244353) * N1 % (998244353));
      }
    } else {
      long long v;
      cin >> v;
      long long ans = addall + add[v];
      while (v != 0) {
        ans += bit.ask(id[head[v]], id[v]);
        v = head[v];
        ans = (ans + add[p[v]] * (N - sz[v]) % (998244353) * N1) % (998244353);
        v = p[v];
      }
      cout << (ans + (998244353)) % (998244353) << "\n";
    }
  }
  return 0;
}