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 namespace std::chrono;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
mt19937 rng(steady_clock::now().time_since_epoch().count());
void usaco(string filename) {
#pragma message("be careful, freopen may be wrong")
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
const long double pi = 3.14159265358979323846;
const long long mod = 998244353;
namespace modop {
long long madd(long long a, long long b) { return (a + b) % mod; }
long long msub(long long a, long long b) {
  return (((a - b) % mod) + mod) % mod;
}
long long mmul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long mpow(long long base, long long exp) {
  long long res = 1;
  while (exp) {
    if (exp % 2 == 1) {
      res = (res * base) % mod;
    }
    exp >>= 1;
    base = (base * base) % mod;
  }
  return res;
}
long long minv(long long base) { return mpow(base, mod - 2); }
long long mdiv(long long a, long long b) { return mmul(a, minv(b)); }
}  // namespace modop
using namespace modop;
struct bit {
  int n;
  vector<long long> a;
  void init(int second) {
    a = vector<long long>(second + 1, 0);
    n = second;
  }
  void add(int loc, long long x) {
    if (loc <= 0) return;
    while (loc <= n) {
      a[loc] += x;
      loc += loc & (-loc);
    }
  }
  void radd(int l, int r, long long x) {
    add(l + 1, x);
    add(r + 2, -x);
  }
  long long query(int loc) {
    ++loc;
    long long sum = 0;
    while (loc > 0) {
      sum += a[loc];
      loc -= loc & (-loc);
    }
    return sum;
  }
#pragma message("BITs are 1-indexed, be careful")
};
struct lca_lift {
  const int lg = 18;
  int n;
  vector<int> depth;
  vector<vector<int> > edges;
  vector<vector<int> > lift;
  void init(int sz) {
    n = sz;
    depth = vector<int>(n);
    edges = vector<vector<int> >(n, vector<int>());
    lift = vector<vector<int> >(n, vector<int>(lg, -1));
  }
  void edge(int a, int b) {
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  void attach(int node_to_attach, int node_to_attach_to) {
    int a = node_to_attach, b = node_to_attach_to;
    edge(a, b);
    lift[a][0] = b;
    for (int i = 1; i < lg; i++) {
      if (lift[a][i - 1] == -1) break;
      lift[a][i] = lift[lift[a][i - 1]][i - 1];
    }
    depth[a] = depth[b] + 1;
  }
  void init_lift() {
    depth[0] = 0;
    dfs(0, -1);
  }
  void dfs(int v, int par) {
    lift[v][0] = par;
    for (int i = 1; i < lg; i++) {
      if (lift[v][i - 1] == -1) break;
      lift[v][i] = lift[lift[v][i - 1]][i - 1];
    }
    for (int x : edges[v]) {
      if (x != par) {
        depth[x] = depth[v] + 1;
        dfs(x, v);
      }
    }
  }
  int get(int v, int k) {
    for (int i = lg - 1; i >= 0; i--) {
      if ((1 << i) <= k) {
        v = lift[v][i];
        k -= (1 << i);
      }
    }
    return v;
  }
  int get_lca(int a, int b) {
    if (depth[a] < depth[b]) swap(a, b);
    int d = depth[a] - depth[b];
    int v = get(a, d);
    if (v == b) {
      return v;
    } else {
      for (int i = lg - 1; i >= 0; i--) {
        if (lift[v][i] != lift[b][i]) {
          v = lift[v][i];
          b = lift[b][i];
        }
      }
      return lift[b][0];
    }
  }
};
long long n, m, k, q, Q, T, l, r, x, y, z;
long long a[1000005];
long long b[1000005];
long long c[1000005];
string second, t;
long long ans = 0;
vector<int> edges[150005], child[150005];
long long val[150005], sz[150005];
int st[150005], en[150005];
long long tot = 0;
int TIME = -1;
const long long C = 400;
vector<int> spec;
bit sumt;
lca_lift lca;
void findc(int v, int p = -1) {
  sz[v] = 1;
  st[v] = ++TIME;
  for (int x : edges[v]) {
    if (x != p) {
      child[v].push_back(x);
      findc(x, v);
      sz[v] += sz[x];
    }
  }
  en[v] = TIME;
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  };
  cin >> n >> Q;
  memset(val, 0, sizeof(val));
  long long MN = minv(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    edges[a].push_back(b);
    edges[b].push_back(a);
  };
  lca.init(n);
  for (long long i = 0; i < n; i++)
    for (int x : edges[i])
      if (i > x) lca.edge(i, x);
  findc(0);
  memset(a, 0, sizeof(a));
  sumt.init(n);
  lca.init_lift();
  for (long long i = 0; i < n; i++)
    if (edges[i].size() >= C) spec.push_back(i);
  for (long long q = 0; q < Q; q++) {
    int type;
    cin >> type;
    if (type == 1) {
      cin >> x >> y;
      --x;
      if (edges[x].size() < C) {
        tot = (tot + sz[x] * y) % mod;
        for (int z : child[x]) {
          sumt.radd(st[z], en[z], (y * (n - sz[z])) % mod);
        }
        sumt.radd(st[x], en[x], msub(0, (y * sz[x]) % mod));
      }
      val[x] = (val[x] + y) % mod;
    } else {
      cin >> x;
      --x;
      long long res = sumt.query(st[x]) % mod;
      res = (res + val[x] * n) % mod;
      res = (res + tot) % mod;
      for (int y : spec) {
        if (y == x) continue;
        int v = lca.get_lca(x, y);
        if (v != y) {
          res = (res + val[y] * sz[y]) % mod;
        } else {
          int dist = lca.depth[x] - lca.depth[y];
          v = lca.get(x, dist - 1);
          res = (res + val[y] * (n - sz[v])) % mod;
        }
      }
      cout << (res * MN) % mod << '\n';
    }
  }
}