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 long long N = 2e5 + 10;
const long long mod = 998244353;
long long fen[N];
void upd(long long x, long long val) {
  for (long long i = x + 1; i < N; i += i & -i) {
    fen[i] += val;
  }
}
long long query(long long x) {
  long long res = 0;
  for (long long i = x + 1; i; i -= i & -i) {
    res += fen[i];
  }
  return res;
}
long long TIME = 0;
long long st[N], ed[N];
long long par[N];
long long hv[N];
long long sz[N];
vector<long long> g[N];
long long e[N];
long long head[N];
void dfs(long long u, long long p) {
  st[u] = TIME++;
  par[u] = p;
  sz[u] = 1;
  for (long long v : g[u]) {
    if (v != p) {
      dfs(v, u);
      sz[u] += sz[v];
      if (hv[u] == -1 || sz[hv[u]] < sz[v]) {
        hv[u] = v;
      }
    }
  }
  ed[u] = TIME;
}
long long powM(long long a, long long b) {
  long long ans = 1;
  for (; b; b = b / 2, a = a * a % mod)
    if (b % 2) ans = ans * a % mod;
  return ans;
}
signed main() {
  memset(hv, -1, sizeof(hv));
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, 0);
  for (long long i = 0; i < n; i++) {
    if (!i || hv[par[i]] ^ i) {
      for (long long j = i; ~j; j = hv[j]) {
        head[j] = i;
      }
    }
  }
  long long inv_n = powM(n, mod - 2);
  for (long long i = 1; i <= q; i++) {
    long long qt, node;
    cin >> qt >> node;
    node--;
    if (qt == 1) {
      long long val;
      cin >> val;
      e[node] += val;
      upd(0, val * sz[node]);
      upd(st[node], -val * sz[node]);
      upd(ed[node], val * sz[node]);
      if (~hv[node]) {
        upd(st[hv[node]], val * (n - sz[hv[node]]));
        upd(ed[hv[node]], -val * (n - sz[hv[node]]));
      }
    } else {
      long long ans = (e[node] * n + query(st[node])) % mod;
      for (long long j = head[node]; j; j = head[par[j]]) {
        ans = (ans + e[par[j]] * (n - sz[j])) % mod;
      }
      ans = ans * inv_n % mod;
      cout << ans << endl;
    }
  }
}