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 = 150 * 1000 + 5, mod = 998244353;
long long n, q, t;
vector<long long> adj[N];
long long seg[N * 4];
long long bg[N];
long long sz[N];
long long st[N];
long long ft[N];
long long dp[N];
long long par[N];
long long a[N];
long long po(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long ans = po(x, y / 2);
  ans *= ans;
  ans %= mod;
  if (y % 2) {
    ans *= x;
    ans %= mod;
  }
  return ans;
}
void dfs1(long long v) {
  st[v] = t;
  t++;
  sz[v] = 1;
  for (auto u : adj[v]) {
    if (u != par[v]) {
      par[u] = v;
      dfs1(u);
      sz[v] += sz[u];
      if (sz[u] > sz[bg[v]]) {
        bg[v] = u;
      }
    }
  }
  ft[v] = t;
}
void dfs2(long long v) {
  for (auto u : adj[v]) {
    if (u != par[v]) {
      if (u == bg[v]) {
        dp[u] = dp[v];
      } else {
        dp[u] = u;
      }
      dfs2(u);
    }
  }
}
void update(long long l, long long r, long long x, long long id = 1,
            long long s = 0, long long e = n) {
  if (r <= s || e <= l) {
    return;
  }
  if (l <= s && e <= r) {
    seg[id] += x;
    seg[id] %= mod;
    return;
  }
  long long md = (s + e) / 2;
  update(l, r, x, id * 2, s, md);
  update(l, r, x, id * 2 + 1, md, e);
}
long long get(long long x, long long id = 1, long long s = 0, long long e = n) {
  if (e <= x || x < s) {
    return 0;
  }
  if (e - s == 1) {
    return seg[id];
  }
  long long md = (s + e) / 2;
  long long ans = seg[id];
  ans += get(x, id * 2, s, md);
  ans %= mod;
  ans += get(x, id * 2 + 1, md, e);
  ans %= mod;
  return ans;
}
int32_t main() {
  cin >> n >> q;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs1(1);
  dp[1] = 1;
  dfs2(1);
  long long np = po(n, mod - 2);
  for (long long i = 0; i < q; i++) {
    long long qt, v;
    cin >> qt >> v;
    if (qt == 1) {
      long long d;
      cin >> d;
      if (bg[v]) {
        long long u = bg[v];
        update(0, st[v], 1ll * sz[v] * d);
        update(ft[v], n, 1ll * sz[v] * d);
        update(st[u], ft[u], 1ll * (n - sz[u]) * d);
        update(st[v], st[v] + 1, 1ll * n * d);
      } else {
        update(0, n, d);
        update(st[v], st[v] + 1, 1ll * (n - 1) * d);
      }
      a[v] += d;
    } else {
      long long ans = get(st[v]);
      v = dp[v];
      while (v != 1) {
        ans += 1ll * a[par[v]] * (n - sz[v]);
        ans %= mod;
        v = dp[par[v]];
      }
      cout << (1ll * np * ans) % mod << endl;
    }
  }
  return 0;
}