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 MAX = 150004;
const long long mod = 998244353;
struct edge {
  long long next, to;
} h[MAX << 1];
long long head[MAX], cnt, cnt_dfsn;
long long n, q, ans, inv;
inline void add(long long x, long long y) {
  h[++cnt] = edge{head[x], y};
  head[x] = cnt;
  h[++cnt] = edge{head[y], x};
  head[y] = cnt;
}
long long dfsn[MAX], size[MAX], son[MAX], fa[MAX], top[MAX], w[MAX];
long long tree[MAX];
inline long long lowbit(long long x) { return x & (-x); }
inline void ad(long long x, long long y) {
  y %= mod;
  for (long long i = x; i <= n; i += lowbit(i)) tree[i] = (tree[i] + y) % mod;
}
inline long long sum(long long x) {
  long long ret = 0;
  for (long long i = x; i; i -= lowbit(i)) ret = (ret + tree[i]) % mod;
  return ret;
}
inline long long quick(long long x, long long y) {
  long long ret = 1ll;
  while (y) {
    if (y & 1) ret = (ret * x) % mod;
    y >>= 1;
    x = (x * x) % mod;
  }
  return ret;
}
void dfs1(long long now, long long father) {
  fa[now] = father;
  size[now] = 1;
  for (long long i = head[now]; i; i = h[i].next) {
    long long aim = h[i].to;
    if (aim != father) {
      dfs1(aim, now);
      size[now] += size[aim];
      if (size[aim] > size[son[now]]) son[now] = aim;
    }
  }
}
void dfs2(long long now, long long start) {
  top[now] = start;
  dfsn[now] = ++cnt_dfsn;
  if (son[now]) dfs2(son[now], start);
  for (long long i = head[now]; i; i = h[i].next) {
    long long aim = h[i].to;
    if (aim != fa[now] && aim != son[now]) {
      dfs2(aim, aim);
    }
  }
}
signed main() {
  scanf("%lld %lld", &n, &q);
  inv = quick(n, mod - 2);
  for (long long x, y, i = 1; i < n; i++) scanf("%lld %lld", &x, &y), add(x, y);
  dfs1(1, 1), dfs2(1, 1);
  fa[1] = 0;
  for (long long op, x, y, i = 1; i <= q; i++) {
    scanf("%lld", &op);
    if (op == 1) {
      scanf("%lld %lld", &x, &y);
      w[x] = (w[x] + y) % mod;
      ad(1, size[x] * y % mod);
      ad(dfsn[x], (mod - size[x] * y % mod) % mod),
          ad(dfsn[x] + size[x], size[x] * y % mod);
      if (son[x])
        ad(dfsn[son[x]], (n - size[son[x]]) * y % mod),
            ad(dfsn[son[x]] + size[son[x]], mod - (n - size[son[x]]) * y % mod);
    } else {
      scanf("%lld", &x);
      ans = (sum(dfsn[x]) + n * w[x] % mod) % mod;
      for (long long j = top[x]; fa[j]; j = top[fa[j]])
        ans = (ans + (n - size[j]) * w[fa[j]] % mod) % mod;
      printf("%lld\n", (ans * inv) % mod);
    }
  }
  return 0;
}