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;
inline int read() {
  int x = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * w;
}
int n, q, hed[200005], cnt, f[200005];
int xu[200005], ed[200005], num, son[200005], tp[200005];
long long sz[200005], tag[200005];
long long mod = 998244353;
struct node {
  int to, next;
} a[200005 << 1];
struct tree {
  int l, r;
  long long s, tag;
} tr[800010];
long long ksm(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
inline void add(int from, int to) {
  a[++cnt].to = to;
  a[cnt].next = hed[from];
  hed[from] = cnt;
}
inline void dfs(int x) {
  sz[x] = 1;
  for (int i = hed[x]; i; i = a[i].next) {
    int y = a[i].to;
    if (y == f[x]) continue;
    f[y] = x;
    dfs(y);
    if (sz[y] > sz[son[x]]) son[x] = y;
    sz[x] += sz[y];
  }
}
inline void build(int rt, int l, int r) {
  tr[rt].l = l;
  tr[rt].r = r;
  tr[rt].s = 0;
  tr[rt].tag = 0;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
}
inline void dfs1(int x, int top) {
  xu[x] = ++num;
  tp[x] = top;
  if (son[x]) {
    dfs1(son[x], top);
    for (int i = hed[x]; i; i = a[i].next) {
      int y = a[i].to;
      if (y == f[x] || y == son[x]) continue;
      dfs1(y, y);
    }
  }
  ed[x] = num;
}
inline void pushdown(int rt) {
  if (tr[rt].tag) {
    tr[rt << 1].tag = (tr[rt << 1].tag + tr[rt].tag) % mod;
    tr[rt << 1 | 1].tag = (tr[rt << 1 | 1].tag + tr[rt].tag) % mod;
    tr[rt << 1].s = (tr[rt << 1].s +
                     tr[rt].tag * (tr[rt << 1].r - tr[rt << 1].l + 1) % mod) %
                    mod;
    tr[rt << 1 | 1].s =
        (tr[rt << 1 | 1].s +
         tr[rt].tag * (tr[rt << 1 | 1].r - tr[rt << 1 | 1].l + 1) % mod) %
        mod;
    tr[rt].tag = 0;
  }
}
inline void update(int rt, int le, int ri, long long x) {
  int l = tr[rt].l, r = tr[rt].r;
  if (le <= l && r <= ri) {
    tr[rt].tag = (tr[rt].tag + x) % mod;
    tr[rt].s = (tr[rt].s + x * (r - l + 1) % mod) % mod;
    return;
  }
  pushdown(rt);
  int mid = (l + r) >> 1;
  if (le <= mid) update(rt << 1, le, ri, x);
  if (mid + 1 <= ri) update(rt << 1 | 1, le, ri, x);
  tr[rt].s = (tr[rt << 1].s + tr[rt << 1 | 1].s) % mod;
}
inline long long query(int rt, int le, int ri) {
  int l = tr[rt].l, r = tr[rt].r;
  if (le <= l && r <= ri) return tr[rt].s;
  pushdown(rt);
  int mid = (l + r) >> 1;
  long long res = 0;
  if (le <= mid) res = query(rt << 1, le, ri);
  if (mid + 1 <= ri) res = (res + query(rt << 1 | 1, le, ri)) % mod;
  return res;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n - 1; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1);
  build(1, 1, n);
  dfs1(1, 1);
  for (int i = 1; i <= q; ++i) {
    int opt, x;
    long long y = 0;
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%lld", &x, &y);
      if (xu[x] - 1 >= 1) update(1, 1, xu[x] - 1, sz[x] * y % mod);
      if (xu[x] + sz[x] <= n) update(1, xu[x] + sz[x], n, sz[x] * y % mod);
      if (sz[x] > 1)
        update(1, xu[x] + 1, xu[x] + sz[x] - 1, (n - sz[x]) * y % mod);
      if (son[x])
        update(1, xu[son[x]], ed[son[x]], (sz[x] - sz[son[x]]) * y % mod);
      tag[x] = (tag[x] + y) % mod;
    } else {
      scanf("%d", &x);
      long long ans = tag[x] * n % mod;
      while (x) {
        ans = (ans + query(1, xu[x], xu[x])) % mod;
        if (f[tp[x]])
          ans = (ans - query(1, xu[f[tp[x]]], xu[f[tp[x]]]) + mod) % mod;
        ans = (ans + tag[f[tp[x]]] * (sz[f[tp[x]]] - sz[tp[x]]) % mod) % mod;
        x = f[tp[x]];
      }
      ans = ans * ksm(n, mod - 2) % mod;
      printf("%lld\n", ans);
    }
  }
  return 0;
}