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>
const int mod = 998244353;
const int MAXN = 1.5e5 + 5;
template <typename _T>
void read(_T &x) {
  x = 0;
  char s = getchar();
  int f = 1;
  while (s < '0' || '9' < s) {
    f = 1;
    if (s == '-') f = -1;
    s = getchar();
  }
  while ('0' <= s && s <= '9') {
    x = (x << 3) + (x << 1) + (s - '0'), s = getchar();
  }
  x *= f;
}
template <typename _T>
void write(_T x) {
  if (x < 0) putchar('-'), x = -x;
  if (9 < x) write(x / 10);
  putchar(x % 10 + '0');
}
struct Edge {
  int to, nxt;
} Graph[MAXN << 1];
int su[MAXN << 2], BIT[MAXN];
int siz[MAXN], heavy[MAXN], LPos[MAXN], LTail[MAXN], bel[MAXN], dep[MAXN],
    seq[MAXN];
int head[MAXN], coe[MAXN], fath[MAXN], only[MAXN];
int N, Q, cnt, tot, ID, inv;
int Qkpow(int, int);
inline int Mul(long long x, int v) { return x * v % mod; }
inline int Inv(const int a) { return Qkpow(a, mod - 2); }
inline int Sub(int x, int v) { return (x -= v) < 0 ? x + mod : x; }
inline int Add(int x, int v) { return (x += v) >= mod ? x - mod : x; }
inline void Upt(const int x) { su[x] = Add(su[x << 1], su[x << 1 | 1]); }
void Update(int x, int v) {
  for (; x <= N; x += (x & (-x))) BIT[x] = Add(BIT[x], v);
}
int Get(int x) {
  int ret = 0;
  for (; x; x -= (x & (-x))) ret = Add(ret, BIT[x]);
  return ret;
}
void Update(int l, int r, int v) {
  if (l <= r) Update(l, v), Update(r + 1, mod - v);
}
int Qkpow(int base, int indx) {
  int ret = 1;
  while (indx) {
    if (indx & 1) ret = Mul(ret, base);
    base = Mul(base, base), indx >>= 1;
  }
  return ret;
}
void AddEdge(const int from, const int to) {
  Graph[++cnt].to = to, Graph[cnt].nxt = head[from];
  head[from] = cnt;
}
void DFS1(const int u, const int fa) {
  fath[u] = fa;
  dep[u] = dep[fa] + 1, siz[u] = 1, heavy[u] = -1;
  for (int i = head[u], v; i; i = Graph[i].nxt)
    if ((v = Graph[i].to) ^ fa) {
      DFS1(v, u), siz[u] += siz[v];
      if (heavy[u] == -1 || siz[heavy[u]] < siz[v]) heavy[u] = v;
    }
  coe[u] = Mul(N - siz[u], inv);
}
void DFS2(const int u, const int fa) {
  seq[LPos[u] = ++ID] = u;
  if (~heavy[u])
    DFS2(heavy[u], u), bel[u] = bel[heavy[u]];
  else
    bel[u] = ++tot;
  LTail[bel[u]] = u;
  for (int i = head[u], v; i; i = Graph[i].nxt)
    if ((v = Graph[i].to) ^ fa && v ^ heavy[u]) DFS2(v, u);
}
void Init() {
  inv = Inv(N);
  DFS1(1, 0), DFS2(1, 0);
}
void Update(const int x, const int l, const int r, const int p, const int v) {
  if (l == r) {
    su[x] = Add(su[x], v);
    return;
  }
  int mid = l + r >> 1;
  if (p <= mid)
    Update(x << 1, l, mid, p, v);
  else
    Update(x << 1 | 1, mid + 1, r, p, v);
  Upt(x);
}
int Query(const int x, const int l, const int r, const int segL,
          const int segR) {
  if (segL <= l && r <= segR) return su[x];
  int mid = l + r >> 1, ret = 0;
  if (segL <= mid) ret = Add(ret, Query(x << 1, l, mid, segL, segR));
  if (mid < segR) ret = Add(ret, Query(x << 1 | 1, mid + 1, r, segL, segR));
  return ret;
}
void Change(const int u, const int val) {
  int nw = Mul(val, Mul(siz[u], inv));
  Update(1, LPos[u] - 1, nw);
  Update(LPos[u] + siz[u], N, nw);
  only[u] = Add(only[u], val);
  if (~heavy[u]) Update(1, 1, N, LPos[heavy[u]], Mul(coe[heavy[u]], val));
}
int Query(int u) {
  int ret = Add(Get(LPos[u]), only[u]);
  while (u) {
    ret = Add(ret, Query(1, 1, N, LPos[LTail[bel[u]]], LPos[u]));
    u = LTail[bel[u]];
    if (fath[u]) ret = Add(ret, Mul(coe[u], only[fath[u]]));
    u = fath[u];
  }
  return ret;
}
int main() {
  read(N), read(Q);
  for (int i = 1, a, b; i < N; i++)
    read(a), read(b), AddEdge(a, b), AddEdge(b, a);
  Init();
  for (int opt, u, d; Q; --Q) {
    read(opt), read(u);
    if (opt == 1)
      read(d), Change(u, d);
    else
      write(Query(u)), putchar('\n');
  }
  return 0;
}