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 inf = (int)1e9 + 7;
const int N = (int)4e5 + 5;
const int mod = 998244353;
int binpow(int x, int y) {
  if (y == 0) {
    return 1;
  }
  int z = binpow(x, y >> 1);
  z = z * 1ll * z % mod;
  if (y & 1) {
    z = z * 1ll * x % mod;
  }
  return z;
}
int n, q;
int sub[N], curans[N], sub2[N], lg[N];
int h[N], tour[N], l[N], r[N], par[N], T;
int mn[20][N];
vector<int> g[N], g2[N];
void dfs(int v, int p) {
  sub[v] = 1;
  h[v] = h[p] + 1;
  par[v] = p;
  l[v] = ++T;
  tour[T] = v;
  for (int i : g[v]) {
    if (i == p) continue;
    g2[v].push_back(i);
    dfs(i, v);
    sub[v] += sub[i];
  }
  r[v] = T;
}
inline int ancestor(int a, int b) { return l[a] <= l[b] && r[b] <= r[a]; }
int inv;
int szz;
int id[N];
pair<int, int> qrs[N];
inline int best(int a, int b) {
  return h[a] == h[b] ? (l[a] > l[b] ? a : b) : (h[a] < h[b] ? a : b);
}
inline int getMin(int l, int r) {
  int z = lg[r - l + 1];
  return best(mn[z][l], mn[z][r - (1 << z) + 1]);
}
inline int findSecond(int a, int b) {
  int L = l[a];
  int R = l[b];
  return getMin(L + 1, R);
}
inline void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
}
int f[N];
int u[N];
void refresh() {
  for (int i = 1; i <= n; ++i) f[i] = 0;
  for (int j = 1; j <= szz; ++j) {
    pair<int, int> i = qrs[j];
    u[i.first] = 0;
    add(f[1], sub[i.first] * 1ll * i.second % mod);
    add(curans[i.first], (n - sub[i.first]) * 1ll * i.second % mod);
    for (int j : g2[i.first]) {
      add(f[j], (sub[i.first] * 1ll * i.second % mod) * -1);
      add(f[j], (n - sub[j]) * 1ll * i.second % mod);
    }
  }
  for (int j = 1; j <= T; ++j) {
    add(f[tour[j]], f[par[tour[j]]]);
    add(curans[tour[j]], f[tour[j]]);
  }
  szz = 0;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= T; ++i) {
    mn[0][i] = tour[i];
    if (i > 1) {
      lg[i] = lg[i / 2] + 1;
    }
  }
  for (int j = 1; j < 20; ++j) {
    for (int i = 1; i + (1 << j) - 1 <= T; ++i) {
      mn[j][i] = best(mn[j - 1][i], mn[j - 1][i + (1 << (j - 1))]);
    }
  }
  inv = binpow(n, mod - 2);
  for (int i = 1; i <= q; ++i) {
    if (szz > 650) {
      refresh();
    }
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      int v, d;
      scanf("%d %d", &v, &d);
      if (u[v]) {
        add(qrs[id[v]].second, d);
      } else {
        u[v] = 1;
        qrs[++szz] = make_pair(v, d);
        id[v] = szz;
      }
    } else {
      int v;
      scanf("%d", &v);
      int ans = curans[v];
      for (int j = 1; j <= szz; ++j) {
        pair<int, int> i = qrs[j];
        if (i.first == v) {
          ans += i.second * 1ll * n % mod;
          if (ans >= mod) ans -= mod;
          continue;
        }
        if (ancestor(i.first, v)) {
          int vv = findSecond(i.first, v);
          int toval = i.second * 1ll * (n - sub[vv]) % mod;
          ans += toval;
          if (ans >= mod) {
            ans -= mod;
          }
        } else {
          int toval = sub[i.first] * 1ll * i.second % mod;
          ans += toval;
          if (ans >= mod) {
            ans -= mod;
          }
        }
      }
      printf("%d\n", ans * 1ll * inv % mod);
    }
  }
  return 0;
}