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 uint = unsigned int;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &item) {
  out << '(' << item.first << ", " << item.second << ')';
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (const auto &item : v) out << item << ' ';
  return out;
}
const int N = 150010;
const int MOD = 998244353;
int k, n, invn;
int tin[N], tout[N], bit[N], store[N], nxt[N], heavyChild[N], fth[N], sz[N];
vector<int> adj[N];
int dfs(int v, int p = -1) {
  int heavy = -1, maxSz = -1;
  sz[v] = 1;
  tin[v] = k++;
  for (auto u : adj[v])
    if (u != p) {
      int aux = dfs(u, v);
      fth[u] = v;
      sz[v] += aux;
      if (aux > maxSz) maxSz = aux, heavy = u;
    }
  heavyChild[v] = heavy;
  tout[v] = k;
  return sz[v];
}
void dfsLight(int v, int lightV = -1, int p = -1) {
  nxt[v] = lightV;
  for (auto u : adj[v])
    if (u != p) {
      dfsLight(u, heavyChild[v] == u ? lightV : u, v);
    }
}
int pw(int base, int exp, int mod) {
  int res;
  for (res = 1; exp; exp >>= 1) {
    if (exp & 1) res = (1LL * res * base) % mod;
    base = (1LL * base * base) % mod;
  }
  return res;
}
int modInv(int base, int mod) { return pw(base, mod - 2, mod); }
void add(int &x, int y) {
  if ((x += y) >= MOD) x -= MOD;
}
int sum(int x, int y) { return (x + y < MOD ? x + y : x + y - MOD); }
void update(int p, int val) {
  for (; p; p &= p - 1) add(bit[p], val);
}
void update(int l, int r, int val) {
  if (val) {
    update(l - 1, MOD - val);
    update(r, val);
  }
}
int query(int p) {
  int res = 0;
  for (; p < N; p += p & (-p)) add(res, bit[p]);
  return res;
}
void process(int v, int d) {
  if (heavyChild[v] > 0) {
    int u = heavyChild[v];
    int prob = (1LL * (n - sz[u]) * invn) % MOD;
    update(tin[u], tout[u] - 1, (1LL * prob * d) % MOD);
  }
  int prob = (1LL * sz[v] * invn) % MOD;
  int x = (1LL * prob * d) % MOD;
  if (x) {
    update(1, n, x);
    update(tin[v], tout[v] - 1, MOD - x);
  }
  add(store[v], d);
}
int solve(int v) {
  int res = sum(store[v], query(tin[v]));
  while (nxt[v] > 0) {
    v = nxt[v];
    int prob = (1LL * (n - sz[v]) * invn) % MOD;
    int x = (1LL * prob * store[fth[v]]) % MOD;
    add(res, x);
    v = nxt[fth[v]];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  int q, a, b;
  cin >> n >> q;
  invn = modInv(n, MOD);
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  k = 1;
  dfs(1);
  dfsLight(1);
  int type, v, d;
  while (q--) {
    cin >> type;
    if (type == 1) {
      cin >> v >> d;
      process(v, d);
    } else {
      cin >> v;
      cout << solve(v) << '\n';
    }
  }
  return 0;
}