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;
using ll = long long;
const long long mod = 998244353, maxn = 1e5 + 5e4 + 5;
void add(long long& x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long res = power(a, b / 2);
  if (b & 1) return 1ll * res * res % mod * a % mod;
  return 1ll * res * res % mod;
}
long long inv(long long a) { return power(a, mod - 2); }
class BIT {
 public:
  long long bit[maxn];
  void upd(long long i, long long val) {
    for (; i < maxn; i += i & -i) add(bit[i], val);
  }
  long long query(long long i) {
    long long res = 0;
    for (; i; i -= i & -i) add(res, bit[i]);
    return res;
  }
  void upd(long long l, long long r, ll val) {
    upd(l, val);
    upd(r + 1, -val);
  }
} ft;
long long N, Q;
vector<long long> adj[maxn];
long long tin[maxn], tout[maxn], lazy[maxn], sz[maxn];
vector<long long> big;
void dfs(long long u, long long p = -1) {
  static long long nTime = 0;
  tin[u] = ++nTime;
  sz[u] = 1;
  if (p != -1) {
    auto it = find(adj[u].begin(), adj[u].end(), p);
    adj[u].erase(it);
  }
  for (long long v : adj[u]) {
    dfs(v, u);
    sz[u] += sz[v];
  }
  tout[u] = nTime;
}
bool isAnc(long long u, long long v) {
  return tin[u] <= tin[v] && tout[v] <= tout[u];
}
long long child(long long u, long long v) {
  long long l = 0, r = adj[u].size() - 1, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    long long a = adj[u][mid];
    if (tin[a] <= tin[v])
      l = mid + 1;
    else
      r = mid - 1;
  }
  return adj[u][r];
}
signed main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> Q;
  for (long long i = 1; i < N; ++i) {
    long long u, v;
    cin >> u >> v;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  dfs(1);
  long long magic = sqrt(N);
  for (long long i = 1; i <= N; ++i) {
    if (adj[i].size() > magic) {
      big.emplace_back(i);
    }
  }
  for (long long i = 1; i <= Q; ++i) {
    long long type, v;
    cin >> type >> v;
    if (type == 1) {
      long long val;
      cin >> val;
      if (adj[v].size() > magic) {
        add(lazy[v], val);
      } else {
        for (long long c : adj[v]) {
          ft.upd(tin[c], tout[c], 1ll * (N - sz[c]) * val % mod);
        }
        ll req = 1ll * sz[v] * val % mod;
        if (tin[v] > 1) ft.upd(1, tin[v] - 1, req);
        if (tout[v] < N) ft.upd(tout[v] + 1, N, req);
        ft.upd(tin[v], tin[v], (ll)val * N % mod);
      }
    } else {
      long long res = ft.query(tin[v]);
      if (adj[v].size() > magic) add(res, 1ll * lazy[v] * N % mod);
      for (long long u : big) {
        if (u == v) continue;
        if (isAnc(u, v)) {
          add(res, (ll)lazy[u] * (N - sz[child(u, v)]) % mod);
        } else {
          add(res, (ll)lazy[u] * sz[u] % mod);
        }
      }
      res = (ll)res * inv(N) % mod;
      cout << res << '\n';
    }
  }
}