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;
using ll = long long;
const int N = 2e5 + 5;
const ll mod = 998244353;
const ll base = 1e7 + 7;
ll m, n, t, k, h[N], d, a[N], tong, ans, sz[N], in[N], out[N], pos[N], heavy[N],
    lazy[N], par[N], fe[N];
vector<ll> kq, adj[N];
void ad(ll& x, ll y) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
}
inline ll pw(ll k, ll n) {
  ll total = 1;
  for (; n; n >>= 1) {
    if (n & 1) total = (total * k) % mod;
    k = (k * k) % mod;
  }
  return total;
}
void dfs(ll u, ll p) {
  in[u] = ++k;
  sz[u] = 1;
  if (p != 0) {
    auto it = find(adj[u].begin(), adj[u].end(), p);
    adj[u].erase(it);
  }
  for (ll v : adj[u]) {
    if (v == p) continue;
    par[v] = u;
    dfs(v, u);
    sz[u] += sz[v];
  }
  out[u] = k;
}
void add(ll id, ll val) {
  for (; id <= n; id += id & -id) ad(fe[id], val);
}
inline ll gets(ll id) {
  ll total = 0;
  for (; id; id -= id & -id) {
    ad(total, fe[id]);
  }
  return total;
}
bool inside(ll i, ll j) { return in[j] <= in[i] && out[i] <= out[j]; }
void up(ll l, ll r, ll val) {
  if (l > r) return;
  add(l, val);
  add(r + 1, -val);
}
bool cmp(ll x, ll y) { return (in[x] < in[y]); }
inline ll findc(ll u, ll v) {
  ll lf = 0, rt = adj[u].size() - 1, mid;
  while (lf <= rt) {
    mid = (lf + rt) / 2;
    if (in[adj[u][mid]] <= in[v])
      lf = mid + 1;
    else
      rt = mid - 1;
  }
  return adj[u][rt];
}
void fastscan(int& x) {
  register int c;
  x = 0;
  while ((c = getchar()) <= 47 || c >= 58)
    ;
  for (; (c > 47 && c < 58); c = getchar()) {
    x = (x << 3) + (x << 1) + c - 48;
  }
}
void sol() {
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    ll x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  ll base = pw(n, mod - 2);
  dfs(1, 0);
  t = sqrt(n);
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() >= t) {
      heavy[i] = 1;
      kq.push_back(i);
    }
  }
  while (m-- > 0) {
    ll x, y;
    int query;
    cin >> query;
    if (query == 1) {
      cin >> x >> y;
      if (heavy[x])
        ad(lazy[x], y);
      else {
        for (ll v : adj[x]) {
          t = n - sz[v];
          up(in[v], out[v], t * y % mod);
        }
        t = sz[x];
        up(1, n, y * t % mod);
        up(in[x], out[x], (mod - y * t % mod));
        up(in[x], in[x], n * y % mod);
      }
    } else {
      cin >> x;
      ans = gets(in[x]);
      if (lazy[x]) ad(ans, lazy[x] * n % mod);
      for (ll v : kq) {
        if (x == v || lazy[v] == 0) continue;
        if (inside(x, v)) {
          t = findc(v, x);
          ad(ans, (n - sz[t]) * lazy[v] % mod);
        } else
          ad(ans, sz[v] * lazy[v] % mod);
      }
      cout << ans * base % mod << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("RGAME"
            ".inp",
            "r")) {
    freopen(
        "RGAME"
        ".inp",
        "r", stdin);
    freopen(
        "RGAME"
        ".out",
        "w", stdout);
  }
  sol();
}