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 <class T = int, class S = T>
T input() {
  T x;
  cin >> (S &)x;
  return x;
}
template <class T>
struct Fenwick {
  void init(int n) {
    t.resize(n + 2);
    memset(t.data(), 0, t.size() * sizeof(T));
  }
  int size() { return t.size() - 2; }
  void add(int x, const T &v) {
    for (; x <= size(); x += x & -x) t[x] += v;
  }
  T sum(int x) { return sum<T, T>(x); }
  template <class R, class M>
  R sum(int x) {
    M ret(t[x]);
    while (x &= x - 1) ret += t[x];
    return ret;
  }
  void addi(int x, int y, const T &v) {
    add(x, v);
    add(y + 1, -v);
  }

 private:
  vector<T> t;
};
template <class T, class V>
struct crtp {
  typedef
      typename conditional<is_const<typename remove_pointer<T>::type>::value,
                           const V *, V *>::type type;
};
template <class D>
struct AdditiveOps {
  D operator+(const D &v) const {
    return D(*(static_cast<typename crtp<decltype(this), D>::type>(this))) += v;
  }
};
template <int M, class T = int, class MT = T>
struct IntM : AdditiveOps<IntM<M, T, MT>> {
  IntM() = default;
  static IntM d(T v) { return IntM(v); }
  static IntM pm(T v) { return IntM(v % M); }

 protected:
  explicit IntM(T v) : v(v) {}

 public:
  IntM &operator+=(IntM i) {
    v += i.v;
    if (v >= M) v -= M;
    return *this;
  }
  IntM operator-() const { return IntM(v ? M - v : 0); }
  IntM &operator*=(const IntM i) {
    v = (MT)v * i.v % M;
    return *this;
  }
  IntM operator*(const IntM i) const { return IntM(*this) *= i; }
  friend istream &operator>>(istream &is, IntM &i) { return is >> i.v; }
  friend ostream &operator<<(ostream &os, const IntM &i) { return os << i.v; }
  IntM inv() const {
    pair<T, T> p(v, M);
    pair<T, T> x(0, 1);
    pair<T, T> y(1, 0);
    while (p.second) {
      MT q = p.first / p.second;
      p = pair<T, T>(p.second, p.first % p.second);
      x = pair<T, T>(x.second - q * x.first, x.first);
      y = pair<T, T>(y.second - q * y.first, y.first);
    }
    return IntM::d(x.second < 0 ? x.second + M : x.second);
  }
  T v;
};
constexpr int M = 998244353;
struct Node {
  vector<Node *> neigh;
  int discTime, finTime;
  int subtreeSize;
  Node *heavyAnc;
  Node *heavyAncChild;
  IntM<M, int, long long> updateVal;
  bool isHeavy(int heavyTreshold) const {
    return neigh.size() >= heavyTreshold;
  }
};
void dfs(Node *v, Node *parent, int &curTime, int heavyTreshold) {
  v->discTime = curTime++;
  v->heavyAnc = nullptr;
  v->heavyAncChild = nullptr;
  if (parent) {
    if (parent->isHeavy(heavyTreshold)) {
      v->heavyAnc = parent;
      v->heavyAncChild = v;
    } else {
      v->heavyAnc = parent->heavyAnc;
      v->heavyAncChild = parent->heavyAncChild;
    }
  }
  int childSubtreesSize = 0;
  for (Node *u : v->neigh) {
    if (u == parent) {
      continue;
    }
    dfs(u, v, curTime, heavyTreshold);
    childSubtreesSize += u->subtreeSize;
  }
  v->subtreeSize = 1 + childSubtreesSize;
  v->updateVal = IntM<M, int, long long>::d(0);
  v->finTime = curTime;
}
struct SubtreeSums {
  void init(Node *root) {
    this->root = root;
    bit.init(root->subtreeSize);
  }
  void addi(Node *v, IntM<M, int, long long> x) {
    bit.addi(v->discTime + 1, v->finTime, x);
  }
  void addiUpEx(Node *v, IntM<M, int, long long> x) {
    addi(root, x);
    addi(v, -x);
  }
  IntM<M, int, long long> sum(Node *v) { return bit.sum(v->discTime + 1); }
  Node *root;
  Fenwick<IntM<M, int, long long>> bit;
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<Node> tree(n);
  for (int i = 0; i < n - 1; ++i) {
    Node *a = &tree[input() - 1];
    Node *b = &tree[input() - 1];
    a->neigh.push_back(b);
    b->neigh.push_back(a);
  }
  int heavyTreshold = (int)sqrt(n);
  int curTime = 0;
  Node *root = &tree[0];
  dfs(root, nullptr, curTime, heavyTreshold);
  SubtreeSums bit;
  bit.init(root);
  IntM<M, int, long long> nInv = IntM<M, int, long long>::d(n).inv();
  while (q--) {
    int tp = input();
    Node *v = &tree[input() - 1];
    if (tp == 1) {
      IntM<M, int, long long> d = IntM<M, int, long long>::pm(input());
      bit.addiUpEx(v, d * IntM<M, int, long long>::d(v->subtreeSize) * nInv);
      v->updateVal += d;
      if (!v->isHeavy(heavyTreshold)) {
        for (Node *c : v->neigh) {
          if (c->discTime < v->discTime) {
            continue;
          }
          bit.addi(c,
                   d * IntM<M, int, long long>::d(n - c->subtreeSize) * nInv);
        }
      }
    } else {
      IntM<M, int, long long> sol = bit.sum(v) + v->updateVal;
      for (Node *u = v; u->heavyAnc; u = u->heavyAnc) {
        sol += u->heavyAnc->updateVal *
               IntM<M, int, long long>::d(n - u->heavyAncChild->subtreeSize) *
               nInv;
      }
      cout << sol << '\n';
    }
  }
  return 0;
}