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 = 150100;
const int mod = 998244353;
vector<int> p[N];
vector<int> szs[N];
vector<pair<int, int> > pr[N];
vector<int> qu[N];
int res[N];
bool vis[N];
int sz[N], ty[N];
int n, q;
void init(int x, int pp = 0) {
  sz[x] = 1;
  szs[x].resize(p[x].size());
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (y == pp) continue;
    init(y, x);
    sz[x] += sz[y];
    szs[x][i] = sz[y];
  }
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (y != pp) continue;
    szs[x][i] = n - sz[x];
  }
}
int getSz(int x, int pp = 0) {
  if (vis[x]) return 0;
  int sz = 1;
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (y == pp) continue;
    sz += getSz(y, x);
  }
  return sz;
}
int getCen(int x, int allSz, int &cen, int pp = 0) {
  if (vis[x]) return 0;
  int sz = 1;
  int mx = 0;
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (y == pp) continue;
    int ch = getCen(y, allSz, cen, x);
    mx = max(mx, ch);
    sz += ch;
  }
  mx = max(mx, allSz - sz);
  if (mx <= allSz / 2) cen = x;
  return sz;
}
int bit[N];
void add(int in, int val) {
  in++;
  while (in < N) {
    bit[in] += val;
    if (bit[in] >= mod) bit[in] -= mod;
    in += in & (-in);
  }
}
int cal(int in) {
  in++;
  int res = 0;
  while (in > 0) {
    res += bit[in];
    if (res >= mod) res -= mod;
    in -= in & (-in);
  }
  return res;
}
void addToBIT(int x, int val, int pp) {
  if (vis[x]) return;
  int sz = 0;
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (y == pp) {
      sz = n - szs[x][i];
      continue;
    }
    addToBIT(y, val, x);
  }
  for (int i = 0; i < pr[x].size(); i++) {
    int64_t d = pr[x][i].second;
    int tim = pr[x][i].first;
    d *= sz;
    d *= val;
    d %= mod;
    if (d < 0) d += mod;
    add(tim, d);
  }
}
void calFromBIT(int x, int64_t val, int pp = 0) {
  if (vis[x]) return;
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (y == pp) {
      continue;
    }
    calFromBIT(y, val, x);
  }
  for (int i = 0; i < qu[x].size(); i++) {
    res[qu[x][i]] += (cal(qu[x][i]) * val) % mod;
    res[qu[x][i]] %= mod;
  }
}
void div(int x) {
  int sz = getSz(x);
  int cen;
  getCen(x, sz, cen);
  vis[cen] = 1;
  x = cen;
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (vis[y]) continue;
    div(y);
  }
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (vis[y]) continue;
    addToBIT(y, 1, x);
  }
  for (int i = 0; i < qu[x].size(); i++) {
    res[qu[x][i]] += cal(qu[x][i]);
    res[qu[x][i]] %= mod;
  }
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (vis[y]) continue;
    addToBIT(y, -1, x);
    calFromBIT(y, 1);
    addToBIT(y, 1, x);
  }
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (vis[y]) continue;
    addToBIT(y, -1, x);
  }
  for (int i = 0; i < pr[x].size(); i++) {
    int64_t d = pr[x][i].second;
    int tim = pr[x][i].first;
    add(tim, d);
  }
  for (int i = 0; i < p[x].size(); i++) {
    int y = p[x][i];
    if (vis[y]) continue;
    calFromBIT(y, n - szs[x][i], x);
  }
  for (int i = 0; i < pr[x].size(); i++) {
    int64_t d = pr[x][i].second;
    int tim = pr[x][i].first;
    d *= -1;
    d %= mod;
    d += mod;
    add(tim, d);
  }
  vis[cen] = 0;
}
int pre[N];
int64_t pw(int64_t x, int64_t p) {
  if (!p) return 1;
  int64_t z = pw(x, p / 2);
  z *= z;
  z %= mod;
  if (p % 2 == 0) return z;
  z *= x;
  z %= mod;
  return z;
}
int main() {
  cin >> n >> q;
  int x, y;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &x, &y);
    p[x].push_back(y);
    p[y].push_back(x);
  }
  int v, d;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &ty[i], &v);
    if (ty[i] == 1) {
      scanf("%d", &d);
      pr[v].push_back({i, d});
      pre[v] += ((int64_t)d * n) % mod;
      if (pre[v] >= mod) pre[v] -= mod;
    } else {
      res[i] += pre[v];
      if (res[i] >= mod) res[i] -= mod;
      qu[v].push_back(i);
    }
  }
  init(1);
  div(1);
  for (int i = 0; i < q; i++) {
    if (ty[i] == 2) {
      res[i] = ((int64_t)res[i] * pw(n, mod - 2)) % mod;
      res[i] %= mod;
      printf("%d\n", res[i]);
    }
  }
  return 0;
}