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>
#pragma GCC optimize("O3")
using namespace std;
const int mod = 998244353;
int sum(int a, int b) {
  int s = a + b;
  if (s >= mod) s -= mod;
  return s;
}
int mult(int a, int b) { return (1LL * a * b) % mod; }
int pw(int a, int b) {
  if (b == 0) return 1;
  if (b & 1) return mult(a, pw(a, b - 1));
  int res = pw(a, b / 2);
  return mult(res, res);
}
int sub(int a, int b) {
  int s = a - b;
  if (s < 0) s += mod;
  return s;
}
const int BLOCK = 400;
const int LG = 19;
const int maxN = 15 * 10000 + 10;
int h[maxN];
int up[LG][maxN];
int n, q;
vector<int> g[maxN];
int tin[maxN];
int tout[maxN];
int timer = 0;
int sz[maxN];
int invn;
int his_batya[maxN];
void dfs(int v, int p) {
  tin[v] = timer++;
  sz[v] = 1;
  for (int to : g[v]) {
    if (to == p) continue;
    h[to] = h[v] + 1;
    his_batya[to] = v;
    up[0][to] = v;
    dfs(to, v);
    sz[v] += sz[to];
  }
  tout[v] = timer++;
}
bool is_anc(int who, int batya) {
  return tin[batya] <= tin[who] && tout[who] <= tout[batya];
}
int parent(int v, int batya) {
  for (int i = LG - 1; i >= 0; i--) {
    if (h[v] - (1 << i) > h[batya]) {
      v = up[i][v];
    }
  }
  return v;
}
long long getval(int u, int v, int d) {
  if (u == v) return 1LL * d * n;
  if (!is_anc(u, v)) {
    int coef = n - sz[v];
    return (1LL * d * (n - coef));
  } else {
    int par = parent(u, v);
    int coef = sz[par];
    return (1LL * d * (n - coef));
  }
}
long long has_now[maxN];
long long coef[maxN];
long long to_add[maxN];
long long cur_val[maxN];
void process(int v, int p) {
  cur_val[v] = cur_val[v] + 1LL * coef[v] * (sz[v]);
  for (int to : g[v]) {
    if (to == p) continue;
    to_add[to] = to_add[v];
    to_add[to] += 1LL * coef[v] * (n - sz[to]);
    cur_val[to] += cur_val[v];
    process(to, v);
  }
}
int tit = 0;
int id[maxN];
vector<int> gg[maxN];
void dfs_start(int v, int p) {
  tit++;
  id[v] = tit;
  for (int to : g[v]) {
    if (to == p) continue;
    dfs_start(to, v);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> q;
  invn = pw(n, mod - 2);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  dfs_start(1, -1);
  for (int i = 1; i <= n; i++) {
    for (int to : g[i]) {
      gg[id[i]].emplace_back(id[to]);
    }
  }
  for (int i = 1; i <= n; i++) {
    swap(gg[i], g[i]);
  }
  dfs(1, -1);
  for (int i = 0; i + 1 < LG; i++) {
    for (int j = 1; j <= n; j++) {
      up[i + 1][j] = up[i][up[i][j]];
    }
  }
  for (int it = 0;; it++) {
    int l = it * BLOCK;
    int r = min(it * BLOCK + BLOCK - 1, q - 1);
    if (l > r) break;
    vector<pair<int, int> > add;
    memset(to_add, 0, sizeof to_add);
    for (int i = l; i <= r; i++) {
      int tp;
      cin >> tp;
      if (tp == 2) {
        int v;
        cin >> v;
        v = id[v];
        long long val = has_now[v];
        for (auto s : add) {
          val += getval(v, s.first, s.second);
        }
        cout << mult(val % mod, invn) << '\n';
      } else {
        int v, d;
        cin >> v >> d;
        v = id[v];
        coef[v] += d;
        add.emplace_back(v, d);
      }
    }
    process(1, -1);
    long long all_sum = 0;
    for (int i = 1; i <= n; i++) all_sum += 1LL * coef[i] * (sz[i]);
    for (int i = 1; i <= n; i++) {
      has_now[i] += all_sum - cur_val[i];
      has_now[i] += 1LL * n * coef[i];
      has_now[i] += to_add[i];
      to_add[i] = 0;
      cur_val[i] = 0;
      coef[i] = 0;
    }
  }
  return 0;
}