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 = 2e5 + 10;
const long long mo = 998244353;
vector<int> vec[maxn];
void addedge(int u, int v) {
  vec[u].push_back(v);
  vec[v].push_back(u);
}
struct SGT {
  long long tree[maxn << 2], lazy[maxn << 2];
  void pushdown(int rt) {
    if (lazy[rt]) {
      lazy[rt << 1] = (lazy[rt << 1] + lazy[rt]) % mo;
      lazy[rt << 1 | 1] = (lazy[rt << 1 | 1] + lazy[rt]) % mo;
      tree[rt << 1] = (tree[rt << 1] + lazy[rt]) % mo;
      tree[rt << 1 | 1] = (tree[rt << 1 | 1] + lazy[rt]) % mo;
      lazy[rt] = 0;
    }
  }
  void build() {
    memset(tree, 0, sizeof(tree));
    memset(lazy, 0, sizeof(lazy));
  }
  void update(int L, int R, int l, int r, long long c, int rt) {
    if (L > R) return;
    if (L <= l && r <= R) {
      tree[rt] = (tree[rt] + c) % mo;
      lazy[rt] = (lazy[rt] + c) % mo;
      return;
    }
    pushdown(rt);
    int mid = (l + r) >> 1;
    if (L <= mid) update(L, R, l, mid, c, rt << 1);
    if (R > mid) update(L, R, mid + 1, r, c, rt << 1 | 1);
  }
  long long query(int L, int l, int r, int rt) {
    if (l == r) {
      return tree[rt];
    }
    pushdown(rt);
    int mid = (l + r) >> 1;
    if (L <= mid)
      return query(L, l, mid, rt << 1);
    else
      return query(L, mid + 1, r, rt << 1 | 1);
  }
} sgt;
int tol;
int fa[maxn], id[maxn], sz[maxn], son[maxn], top[maxn];
long long val[maxn];
void init() {
  tol = 0;
  sgt.build();
  sz[0] = 0;
  memset(val, 0, sizeof(val));
}
void dfs1(int u, int f) {
  id[u] = ++tol;
  fa[u] = f;
  sz[u] = 1;
  son[u] = 0;
  for (int i = 0; i < vec[u].size(); i++) {
    int v = vec[u][i];
    if (v == f) continue;
    dfs1(v, u);
    sz[u] += sz[v];
    if (sz[son[u]] < sz[v]) {
      son[u] = v;
    }
  }
}
void dfs2(int u, int f) {
  top[u] = f;
  for (int i = 0; i < vec[u].size(); i++) {
    int v = vec[u][i];
    if (v == fa[u]) continue;
    if (v == son[u])
      dfs2(v, f);
    else
      dfs2(v, v);
  }
}
long long quick_pow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = (ans * x) % mo;
    y >>= 1;
    x = (x * x) % mo;
  }
  return ans;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
  }
  init();
  dfs1(1, 0);
  dfs2(1, 1);
  long long inv = quick_pow(1LL * n, mo - 2);
  while (q--) {
    int op, v;
    long long d;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%lld", &v, &d);
      val[v] = (val[v] + d) % mo;
      sgt.update(1, id[v] - 1, 1, n, 1LL * sz[v] * d, 1);
      sgt.update(id[v] + sz[v], n, 1, n, 1LL * sz[v] * d, 1);
      sgt.update(id[son[v]], id[son[v]] + sz[son[v]] - 1, 1, n,
                 1LL * (n - sz[son[v]]) * d, 1);
    } else {
      scanf("%d", &v);
      long long ans =
          (sgt.query(id[v], 1, n, 1) + (1LL * n * val[v]) % mo) % mo;
      for (int i = top[v]; fa[i] != 0; i = top[fa[i]]) {
        ans = (ans + val[fa[i]] * (n - sz[i])) % mo;
      }
      ans = (ans * inv) % mo;
      printf("%lld\n", ans);
    }
  }
  return 0;
}