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>
inline void rd(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 - '0' + c, c = getchar();
  x *= f;
}
const int N = 150010, mod = 998244353;
int Pow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = res * (long long)x % mod;
    x = x * (long long)x % mod, y >>= 1;
  }
  return res;
}
int head[N], ecnt;
int sz[N], fa[N];
struct ed {
  int to, next;
} e[N << 1];
void ad(int x, int y) {
  e[++ecnt] = (ed){y, head[x]};
  head[x] = ecnt;
  e[++ecnt] = (ed){x, head[y]};
  head[y] = ecnt;
}
namespace __ {
int *U[N], *D[N];
int dep[N], son[N], top[N];
int dis[N];
int p[N][19];
void dfs1(int u, int last) {
  fa[u] = p[u][0] = last, dis[u] = dis[last] + 1;
  sz[u] = 1;
  for (int j = 1; j <= 18; ++j) p[u][j] = p[p[u][j - 1]][j - 1];
  for (int k = head[u]; k; k = e[k].next) {
    int v = e[k].to;
    if (v == last) continue;
    dfs1(v, u);
    if (dep[v] >= dep[u]) dep[u] = dep[v], son[u] = v;
    sz[u] += sz[v];
  }
  dep[u]++;
}
void dfs2(int u, int tp) {
  top[u] = tp;
  if (son[u]) dfs2(son[u], tp);
  for (int k = head[u]; k; k = e[k].next) {
    int v = e[k].to;
    if (v == p[u][0] || v == son[u]) continue;
    dfs2(v, v);
  }
}
void gao(int u) {
  U[u] = new int[dep[u] + 1];
  D[u] = new int[dep[u] + 1];
  int cur = u;
  for (int i = 0; i <= dep[u]; ++i) D[u][i] = cur, cur = son[cur];
  cur = u;
  for (int i = 0; i <= dep[u]; ++i) U[u][i] = cur, cur = p[cur][0];
}
int Lg[N];
int query(int u, int k) {
  if (dis[u] <= k) return 0;
  if (!k) return u;
  u = p[u][Lg[k]], k -= (1 << Lg[k]);
  int td = dis[u] - dis[top[u]];
  if (td >= k)
    return D[top[u]][td - k];
  else
    return U[top[u]][k - td];
}
}  // namespace __
struct Que {
  int u, d;
};
vector<Que> s;
int n, q;
int delt[N], del[N];
int ans[N];
void push_down(int u, int last) {
  del[u] = (del[last] + del[u]) % mod;
  ans[u] = (ans[u] + del[u]) % mod;
  for (int k = head[u]; k; k = e[k].next) {
    int v = e[k].to;
    if (v == last) continue;
    push_down(v, u);
  }
  del[u] = 0;
}
int get(int v, int u) {
  if (__::dis[v] > __::dis[u]) return sz[v];
  if (__::dis[v] == __::dis[u]) {
    if (v == u) return n;
    return sz[v];
  }
  int t = __::query(u, __::dis[u] - __::dis[v] - 1);
  if (fa[t] == v)
    return n - sz[t];
  else
    return sz[v];
}
int main() {
  rd(n), rd(q);
  int inv = Pow(n, mod - 2);
  for (int i = 1, x, y; i < n; ++i) rd(x), rd(y), ad(x, y);
  __::dfs1(1, 0), __::dfs2(1, 1);
  for (int i = 1; i <= n; ++i)
    if (__::top[i] == i) __::gao(i);
  for (int i = 2; i <= n; ++i) __::Lg[i] = __::Lg[i >> 1] + 1;
  for (int i = 1; i <= q; ++i) {
    int ty, v, d;
    rd(ty), rd(v);
    if (ty == 1) {
      rd(d);
      delt[v] = (delt[v] + d) % mod;
      s.push_back((Que){v, d});
    } else {
      long long t = ans[v];
      for (int j = 0; j < s.size(); ++j)
        t = (t + get(s[j].u, v) * (long long)s[j].d) % mod;
      printf("%lld\n", (t * (long long)inv % mod + mod) % mod);
    }
    if (s.size() > 400) {
      for (int u = 1; u <= n; ++u)
        if (delt[u]) {
          for (int k = head[u]; k; k = e[k].next) {
            int v = e[k].to;
            if (v == fa[u]) continue;
            del[v] = (del[v] + (n - sz[v]) * (long long)delt[u]) % mod;
          }
          del[1] = (del[1] + sz[u] * (long long)delt[u]) % mod;
          del[u] = (del[u] - sz[u] * (long long)delt[u]) % mod;
          ans[u] = (ans[u] + n * (long long)delt[u]) % mod;
          delt[u] = 0;
        }
      push_down(1, 0);
      s.clear();
    }
  }
  return 0;
}