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 ll = long long;
const int MN = 1.5e5 + 1000;
const int MOD = 998244353;
struct intM {
 public:
  int v;
  intM(int v = 0) : v(v) {}
  void in(void) { scanf("%d", &v); }
  operator int() const { return v; }
  intM operator+(int o) const { return intM((v + o) % MOD); }
  intM operator+(intM o) const {
    return v + o.v < MOD ? intM(v + o.v) : intM(v + o.v - MOD);
  }
  void operator+=(intM o) { v = v + o.v < MOD ? v + o.v : v + o.v - MOD; }
  intM operator-(int o) const {
    return v - o < 0 ? intM(v - o + MOD) : intM(v - o);
  }
  intM operator-(intM o) const {
    return v - o.v < 0 ? intM(v - o.v + MOD) : intM(v - o.v);
  }
  intM operator*(int o) const {
    return intM(static_cast<int>(static_cast<ll>(v) * o % MOD));
  }
  intM operator*(intM o) const {
    return intM(static_cast<int>(static_cast<ll>(v) * o.v % MOD));
  }
  void operator*=(intM o) {
    v = static_cast<int>(static_cast<ll>(v) * o.v % MOD);
  }
};
intM pow(intM b, int p) {
  intM r(1);
  for (; p; p >>= 1, b *= b)
    if (p & 1) r *= b;
  return r;
}
intM invN;
int N, Q, R = 1;
std::vector<int> a[MN];
int p[MN], d[MN], s[MN];
int dfs(int n) {
  s[n] = 1;
  for (int x : a[n])
    if (x != p[n]) p[x] = n, d[x] = d[n] + 1, s[n] += dfs(x);
  return s[n];
}
int h[MN], t[MN];
struct HUPD {
 public:
  int w;
  int L;
};
HUPD H[MN];
int C;
void dfs2(int n) {
  for (int x : a[n])
    if (x != p[n]) {
      if (s[x] * 2 >= s[n]) {
        h[n] = x;
        if (!~t[n]) t[n] = n;
        t[x] = t[n];
      }
      dfs2(x);
    }
  if (!~h[n] && ~t[n]) H[t[n]] = {C, d[n] - d[t[n]]}, C += d[n] - d[t[n]];
}
intM bit[MN];
intM all;
intM work[MN], sub[MN];
void bupd(intM* b, int L, int n, intM x) {
  for (; n <= L; n += -n & n) b[n - 1] += x;
}
intM bqry(intM* b, int n) {
  intM r;
  for (; n; n -= -n & n) r += b[n - 1];
  return r;
}
void put(int v, intM d) {
  work[v] += d;
  if (~h[v])
    bupd(bit + H[t[v]].w, H[t[v]].L, ::d[v] - ::d[t[v]] + 1, d * s[h[v]]);
}
void put2(int v, intM d) {
  sub[v] += d;
  if (~t[v] && t[v] != v)
    bupd(bit + H[t[v]].w, H[t[v]].L, ::d[v] - ::d[t[v]], d);
}
intM get(int n) {
  intM r;
  for (; n;)
    if (!~t[n] || t[n] == n)
      r += sub[n] + work[p[n]] * s[n], n = p[n];
    else
      r += bqry(bit + H[t[n]].w, d[n] - d[t[n]]), n = t[n];
  return r;
}
int main(void) {
  memset(h, -1, sizeof(h));
  memset(t, -1, sizeof(t));
  scanf("%d%d", &N, &Q);
  invN = pow(intM(N), MOD - 2);
  for (int i = 1, u, v; i < N; ++i)
    scanf("%d%d", &u, &v), a[u].push_back(v), a[v].push_back(u);
  dfs(R);
  dfs2(R);
  int op, v;
  intM d;
  for (int i = 0; i < Q; ++i) {
    scanf("%d%d", &op, &v);
    if (op == 1) {
      scanf("%d", &d.v);
      if (!d.v) continue;
      put(v, intM(MOD - d.v));
      if (v == R)
        all += d * N;
      else
        all += d * s[v], put2(v, d * (N - s[v]));
    } else
      printf("%d\n", invN * (all + get(v)));
  }
  return 0;
}