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 N = 1.5e5 + 9;
const int mod = 998244353;
int qpow(int n, int k) {
  int ans = 1;
  while (k) {
    if (k & 1) ans = 1LL * ans * n % mod;
    n = 1LL * n * n % mod;
    k >>= 1;
  }
  return ans;
}
int t[N];
void upd(int i, int x) {
  while (i < N) (t[i] += x) %= mod, i += i & -i;
}
void upd(int l, int r, int x) {
  upd(l, x);
  upd(r + 1, mod - x);
}
int query(int i) {
  int ans = 0;
  while (i) (ans += t[i]) %= mod, i -= i & -i;
  return ans;
}
int T, st[N], en[N], sz[N], sum[N];
vector<int> g[N];
void dfs(int u, int pre = 0) {
  if (pre) g[u].erase(find(g[u].begin(), g[u].end(), pre));
  sz[u] = 1;
  for (auto v : g[u]) {
    dfs(v, u);
    sz[u] += sz[v];
  }
  sort(g[u].begin(), g[u].end(), [](int i, int j) { return sz[i] < sz[j]; });
}
vector<pair<int, pair<int, int>>> G[N];
void yo(int u) {
  st[u] = ++T;
  for (auto v : g[u]) yo(v);
  en[u] = T;
  for (int i = 0, j; i < (int)g[u].size(); i = j) {
    for (j = i + 1; j < (int)g[u].size() && sz[g[u][i]] == sz[g[u][j]]; j++)
      ;
    G[u].emplace_back(sz[g[u][i]], make_pair(st[g[u][i]], en[g[u][j - 1]]));
  }
}
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
int32_t main() {
  int n = ({
    int a;
    scanf("%d", &a);
    a;
  }),
      q = ({
        int a;
        scanf("%d", &a);
        a;
      });
  for (int i = 1; i < n; i++) {
    int u = ({
      int a;
      scanf("%d", &a);
      a;
    }),
        v = ({
          int a;
          scanf("%d", &a);
          a;
        });
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  int root = rnd() % n + 1;
  dfs(root);
  yo(root);
  int inv = qpow(n, mod - 2);
  while (q--) {
    int ty = ({
      int a;
      scanf("%d", &a);
      a;
    });
    if (ty == 1) {
      int u = ({
        int a;
        scanf("%d", &a);
        a;
      }),
          k = ({
            int a;
            scanf("%d", &a);
            a;
          });
      int p = 1LL * sz[u] * k % mod;
      upd(1, T, p);
      upd(st[u], en[u], (-p + mod) % mod);
      for (auto x : G[u]) {
        int s = x.first;
        auto p = x.second;
        upd(p.first, p.second, 1LL * (n - s) * k % mod);
      }
      (sum[u] += k) %= mod;
    } else {
      int u = ({
        int a;
        scanf("%d", &a);
        a;
      });
      int ans = query(st[u]);
      ans = 1LL * ans * inv % mod;
      (ans += sum[u]) %= mod;
      printf("%d\n", ans);
    }
  }
  return 0;
}