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 MAXN = 150005;
const int MOD = 998244353;
int n, q, cnt = 1, tim = 0, ofs, inv;
int par[MAXN], siz[MAXN], lo[MAXN], hi[MAXN], ch[MAXN], col[MAXN], head[MAXN],
    fen[MAXN], sol[MAXN];
vector<int> v[MAXN];
int add(int x, int y) {
  x += y;
  if (x >= MOD) return x - MOD;
  return x;
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + MOD;
  return x;
}
int mul(int x, int y) { return (long long)x * y % MOD; }
int pot(int a, int e) {
  int res = 1;
  while (e > 0) {
    if (e & 1) res = mul(res, a);
    a = mul(a, a);
    e /= 2;
  }
  return res;
}
void dfs(int x, int rod) {
  par[x] = rod;
  siz[x] = 1;
  lo[x] = hi[x] = ++tim;
  for (auto sus : v[x]) {
    if (sus == rod) continue;
    dfs(sus, x);
    siz[x] += siz[sus];
    if (siz[sus] > siz[ch[x]]) ch[x] = sus;
    hi[x] = hi[sus];
  }
}
void decompose(int x, bool top) {
  col[x] = cnt;
  if (top) head[cnt] = x;
  if (ch[x] != 0) decompose(ch[x], 0);
  for (auto sus : v[x]) {
    if (sus == ch[x] || sus == par[x]) continue;
    cnt++;
    decompose(sus, 1);
  }
}
void ubaci(int x, int k) {
  for (; x < MAXN; x += x & -x) fen[x] = add(fen[x], k);
}
int upit(int x) {
  int res = 0;
  for (; x > 0; x -= x & -x) res = add(res, fen[x]);
  return res;
}
void update(int x, int d) {
  sol[x] = add(sol[x], d);
  int val = mul(siz[x], d);
  ofs = add(ofs, val);
  ubaci(lo[x], sub(0, val));
  ubaci(hi[x] + 1, val);
  if (ch[x] != 0) {
    val = mul(n - siz[ch[x]], d);
    ubaci(lo[ch[x]], val);
    ubaci(hi[ch[x]] + 1, sub(0, val));
  }
}
int rjesi(int x) {
  int res = add(add(ofs, upit(lo[x])), mul(sol[x], n));
  while (x > 0) {
    x = head[col[x]];
    if (x == 1) break;
    res = add(res, mul(n - siz[x], sol[par[x]]));
    x = par[x];
  }
  return mul(res, inv);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  inv = pot(n, MOD - 2);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1, 0);
  decompose(1, 0);
  for (int i = 0; i < q; i++) {
    int tip;
    cin >> tip;
    if (tip == 1) {
      int x, d;
      cin >> x >> d;
      update(x, d);
    } else {
      int x;
      cin >> x;
      cout << rjesi(x) << '\n';
    }
  }
  return 0;
}