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;
template <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");

 private:
  using ll = long long;
  int v;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend std::istream& operator>>(std::istream& in, modnum& n) {
    ll v_;
    in >> v_;
    n = modnum(v_);
    return in;
  }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  friend modnum inv(const modnum& m) { return m.inv(); }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  friend modnum neg(const modnum& m) { return m.neg(); }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  modnum& operator+=(const modnum& o) {
    v -= MOD - o.v;
    v = (v < 0) ? v + MOD : v;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    v = (v < 0) ? v + MOD : v;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  friend modnum operator/(const modnum& a, const modnum& b) {
    return modnum(a) /= b;
  }
};
const int MAXN = 1.5e5 + 10, MAXQ = 1.5e5 + 10, MOD = 998244353, B = 850;
template <class T>
struct RMQ {
  vector<vector<T>> st;
  vector<int> log;
  T combine(T a, T b) { return min(a, b); }
  void build(int n, vector<T>& a) {
    log = vector<int>(n + 1);
    for (int i = 2; i <= n; ++i) log[i] = log[i >> 1] + 1;
    int k = log[n];
    st = vector<vector<T>>(n, vector<T>(k + 1));
    for (int i = 0; i < n; ++i) st[i][0] = a[i];
    for (int j = 1; j <= k; ++j)
      for (int i = 0; i + (1 << j) - 1 < n; ++i)
        st[i][j] = combine(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  }
  T qry(int l, int r) {
    int k = log[r - l + 1];
    return combine(st[l][k], st[r - (1 << k) + 1][k]);
  }
};
int n, st[MAXN], en[MAXN], depth[MAXN], et[MAXN], par[MAXN], tt = 0;
vector<vector<int>> adj;
RMQ<pair<int, int>> rq;
void init_lc() {
  vector<pair<int, int>> v(n);
  for (int i = 0; i < n; i++) {
    v[i] = {depth[et[i]], -i};
  }
  rq.build(n, v);
}
int almost_anc(int root, int v) {
  if (st[root] <= st[v] && st[v] <= en[root]) {
    auto ans = rq.qry(st[root] + 1, st[v]);
    return et[-ans.second];
  } else
    return par[root];
}
using num = modnum<MOD>;
int q, sub[MAXN];
num add[MAXN], cur[MAXN], prop[MAXN];
vector<pair<int, num>> buffer;
void dfs_init(int c = 0, int p = -1, int d = 0) {
  sub[c] = 1;
  et[st[c] = tt++] = c;
  depth[c] = d, par[c] = p;
  for (auto nxt : adj[c])
    if (nxt != p) {
      dfs_init(nxt, c, d + 1);
      sub[c] += sub[nxt];
    }
  en[c] = tt - 1;
}
void dfs1(int c = 0, int p = -1) {
  cur[c] = add[c] * n;
  for (auto nxt : adj[c])
    if (nxt != p) {
      prop[nxt] += add[c] * (n - sub[nxt]);
      dfs1(nxt, c);
    }
  if (c) {
    prop[c] -= add[c] * sub[c];
    prop[0] += add[c] * sub[c];
  }
}
void dfs2(int c = 0, int p = -1, num sm = 0) {
  sm += prop[c], cur[c] += sm;
  for (auto nxt : adj[c])
    if (nxt != p) dfs2(nxt, c, sm);
}
void bld() {
  fill(prop, prop + n, num(0));
  for (int c = 0; c < n; c++) {
    cur[c] = add[c] * n;
    int p = par[c];
    for (auto nxt : adj[c])
      if (nxt != p) {
        prop[nxt] += add[c] * (n - sub[nxt]);
      }
    if (c) {
      prop[c] -= add[c] * sub[c];
      prop[0] += add[c] * sub[c];
    }
  }
  for (int i = 0; i < n; i++) {
    int c = et[i];
    if (c) prop[c] += prop[par[c]];
    cur[c] += prop[c];
  }
}
void solve() {
  cin >> n >> q;
  adj.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b, --a, --b;
    adj[a].push_back(b), adj[b].push_back(a);
  }
  dfs_init();
  init_lc();
  buffer.reserve(B);
  fill(add, add + n, 0);
  const num IN = num(1) / num(n);
  while (q--) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      int c;
      long long v;
      cin >> c >> v, --c;
      add[c] += v;
      buffer.push_back({c, num(v)});
    } else {
      int c;
      cin >> c, --c;
      num ans = cur[c];
      for (auto v : buffer) {
        if (c == v.first)
          ans += v.second * n;
        else {
          int anc = almost_anc(v.first, c);
          int csub = sub[anc];
          if (anc == par[v.first]) csub = n - sub[v.first];
          ans += v.second * num(n - csub);
        }
      }
      cout << (ans * IN) << '\n';
    }
    if (int(buffer.size()) >= B) {
      bld();
      buffer.clear();
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = 1;
  while (T--) solve();
}